- Example 1 - Solve the equation \(10^{1-x} = 1\)
Solution - \begin{eqnarray} 10^{1-x} &=& 10^0 \\\nonumber 1-x &=& 0\\\nonumber -x &=& -1/\cdot(-1) \\\nonumber x &=& 1\end{eqnarray} - Example 2 - Solve the equation \(100\cdot\left(\frac{1}{2}\right)^{\frac{x}{5}} = 25\)
Solution - \begin{eqnarray}100\cdot\left(\frac{1}{2}\right)^{\frac{x}{5}} &=& 25\\\nonumber 10^2 cdot\left(\frac{1}{2}\right)^{\frac{x}{5}} &=& 5^2/:10^2 \\\nonumber \left(\frac{1}{2}\right)^\frac{x}{5}&=&\left(\frac{1}{2}\right)^2 \\\nonumber \frac{x}{5} &=& 2 /\cdot 5 \\\nonumber x &=& 10\end{eqnarray} - Example 3 - Solve the equation \((0.25)^{1-2x}\cdot \sqrt{64^{1-x}} = \left(\frac{1}{\sqrt{4}}\right)^\frac{x}{2}\)
Solution - \begin{eqnarray}(0.25)^{1-2x}\cdot \sqrt{64^{1-x}} &=& \left(\frac{1}{\sqrt{4}}\right)^\frac{x}{2}\\\nonumber \left(\frac{25}{100}\right)^{1-2x} \cdot \left(2^\frac{6}{2}\right)^{1-x} &=& \left(\frac{1}{2}\right)^\frac{x}{2}\\\nonumber \left(\frac{1}{2}\right)^{2-4x}\cdot 2^{3-3x} &=& \left(\frac{1}{2}\right)^\frac{x}{2}\\\nonumber 2^{-2+4x} \cdot 2^{3-3x} &=& 2^{-\frac{x}{2}}\\\nonumber -2+4x+3-3x &=& -\frac{x}{2}\\\nonumber x+1 &=& -\frac{x}{2}/\cdot 2 \\\nonumber 3x &=& -2 /:3 \\\nonumber x&=& -\frac{2}{3}\end{eqnarray} - Example 4 - Solve the equation \(2^{1+2x} - 9\cdot 2^x + 4 = 0\)
Solution - \begin{eqnarray}2^{1+2x} - 9\cdot 2^x + 4 &=& 0\\\nonumber 2 \cdot 2^{2x} - 9\cdot 2^x + 4 = 0\end{eqnarray} Introducing substitution \(2^x = t,\quad 2^{2x} = t^2\) the equation can be written in the following form \begin{eqnarray}2x^2 - 9t+ 4 &=& 0 \\\nonumber t_{1,2} &=& \frac{9\pm\sqrt{81-4\cdot 4 \cdot 2}}{4} \\\nonumber t_{1,2} &=& \frac{9\pm\sqrt{49}}{4} = \frac{9\pm 7}{4}\\\nonumber t_1 &=& \frac{9+7}{4} = 4 \\\nonumber t_2 &=& \frac{9-7}{4} = \frac{1}{2}\end{eqnarray} Now that the values of variable \(t\) are determined they are used to determine the values of variable \(x\). \begin{eqnarray}2^{x_1} &=& 2^2 \Rightarrow x_1 = 2\\\nonumber 2^{x_2} &=& \frac{1}{2} = 2^{-1}\Rightarrow x_2 = -1\end{eqnarray} - Example 5 - Solve the equation \(4\cdot 3^{x+1} - 3^{x+2} - 3^{x+1} = 72\)
Solution - \begin{eqnarray}4\cdot 3^{x+1} - 3^{x+2} - 3^{x+1} &=& 72 \\\nonumber 12\cdot 3^{x} - 3^x\cdot 3^2 - 3^x \cdot 3^{-1} &=& 72\\\nonumber 3^x\left(12 - 9 -\frac{1}{3}\right) &=& 72 \\\nonumber 3^x\left(\frac{3\cdot 3 - 1}{3}\right) &=& 72\\\nonumber 3^x\left(\frac{8}{3}\right) &=& 72/\cdot \left(\frac{3}{8}\right)\\\nonumber 3^x &=& 27 \\\nonumber x &=& 3\end{eqnarray} - Example 6 - Solve the equation \(2^x + 3^{x-2} = 3^x - 2^{x+1}\)
Solution - \begin{eqnarray}2^x + 3^{x-2} &=& 3^x - 2^{x+1}\\\nonumber 2^x + 2^x \cdot 2 &=& 3^x - 3^x - 3^x \cdot 3^{-2}\\\nonumber 2^x \left(1 +2 \right) &=& 3^x\left(1 - \frac{1}{9}\right)\\\nonumber 3 \cdot 2^x &=& 3^x\left(\frac{9-1}{9}\right)\\\nonumber 3\cdot 2^x &=& 3^x\cdot \frac{8}{9}\\\nonumber 3\cdot 2^x &=& 3^x \cdot 2^3 \cdot 3^{-2}/:(3\cdot 2^3)\\\nonumber 2^{x-3} &=& 3^{x-3}\end{eqnarray} The base numbers in previous equations are not the same but the exponents are. This is possible only if the exponent is equal to zero i.e \(x -3 = 0 \Rightarrow x =3.\) Validation \begin{eqnarray} 2^{3-3} &=& 3^{3-3}\\\nonumber 2^0 &=& 3^0 \\\nonumber 1&=&1\end{eqnarray} - Example 7 - Solve the equation \(10^{x-1} = \frac{1}{10}\)
Solution - \begin{eqnarray}10^{x-1} &=& \frac{1}{10} \\\nonumber 10^{x-1} &=& 10^{-1}\\\nonumber x-1 &=& -1 \\\nonumber x &=& 0\end{eqnarray} - Example 8 - Solve the equation \(2\cdot 2^{2x} + 4^{x+2} -2 \cdot 4^{x-1} =35\)
Solution - \begin{eqnarray}2\cdot 2^{2x} + 4^{x+2} -2 \cdot 4^{x-1} &=&35\\\nonumber 2\cdot 2^{2x} + 2^{2x+4} -2\cdot 2^{2x-2} &=& 35 \\\nonumber 2^{2x}\left(2 + 2^4 - 2^{-1}\right) &=& 35\\\nonumber 2^{2x} \left(\frac{18\cdot 2 - 1}{2}\right) &=& 35\\\nonumber 2^{2x}\left(\frac{35}{2}\right)&=& 35/\cdot \left(\frac{2}{35}\right) \\\nonumber 2^{2x} &=& 2^1\\\nonumber 2x &=& 1 \Rightarrow x = \frac{1}{2}\end{eqnarray} - Example 9 - Solve the equation \(5\cdot 9^{x+1} =15\)
Solution - \begin{eqnarray}5\cdot 9^{x+1} &=& 15/:5 \\\nonumber 3^{2x+2} &=& 3 \\\nonumber 2x+2 &=& 3 \\\nonumber 2x &=& 1/:2 \\\nonumber x &=& \frac{1}{2}\end{eqnarray} - Example 10 - Solve the equation \(100 \cdot 10^x = 0.01\)
Solution - \begin{eqnarray}100 \cdot 10^x &=& 0.01\\\nonumber 10^{x+2} &=& \frac{1}{100} \\\nonumber 10^{x+2}&=& 10^{-2}\\\nonumber x+2 &=& -2 \\\nonumber x &=& -4 \end{eqnarray} - Example 11 - Solve the equation \(10^{x+1} = \frac{1}{10}\)
Solution - \begin{eqnarray}10^{x+1}&=& 10^{-1} \\\nonumber x+ 1 &=& -1 \\\nonumber x &=& -2\end{eqnarray} - Example 12 - Solve the equation \(3\cdot 10^x = 300 \)
Solution - \begin{eqnarray}3\cdot 10^x &=& 300/:3\\\nonumber 10^x &=& 10^2 \\\nonumber x&=& 2\end{eqnarray} - Example 13 - Solve the equation \(2^{2x+1} = \sqrt{8}\)
Solution - \begin{eqnarray}2^{2x+1} &=& \sqrt{8}\\\nonumber 2^{2x+1} &=& 2^{\frac{3}{2}}\\\nonumber 2x+1 &=& \frac{3}{2}\\\nonumber 2x = \frac{3-2}{2}/:2\\\nonumber x &=& \frac{1}{4} \end{eqnarray} - Example 14 - Solve the equation \(64^x - 20\cdot 8^x + 64 = 0\)
Solution - \begin{eqnarray}64^x - 20\cdot 8^x + 64 &=& 0 \\\nonumber 8^{2x} -5\cdot 2^2\cdot 8^x + 2^6 &=& 0 \end{eqnarray} The substitution is introduced in the form \(8^x = t.\) Then the equation can be written as. \begin{eqnarray}t^2 - 20t + 64 &=& 0 \end{eqnarray} \begin{eqnarray}t_{1,2} &=& \frac{20 \pm \sqrt{400 - 4\cdot 64}}{2}\\\nonumber t_{1,2} &=& \frac{20\pm \sqrt{144}}{2} = \frac{20\pm 12}{2}\\\nonumber t_1 &=& \frac{20+12}{2} = \frac{32}{2} = 16 \\\nonumber t_2 &=& \frac{20-12}{2} = \frac{8}{2} = 4\end{eqnarray} Now that values of variable t are determined the next step is to determine the values of variable x using the formula \(8^x = t\) \begin{eqnarray}8^{x_1} &=& t_1 \Rightarrow 8^{x_1} = 16 \Rightarrow 2^{3x_1} = 2^4 \Rightarrow 3x_1 = 4 /:3 \Rightarrow x_1 &=& \frac{4}{3}\\\nonumber 8^{x_2} &=& t_2 \Rightarrow 2^{3x_2} = 2^2 \Rightarrow 3x_2 = 2 \Rightarrow x_2 = \frac{2}{3}\end{eqnarray} - Example 15 - Solve the equation \(3^{x^2-x} - 9 = 0\)
Solution - \begin{eqnarray}3^{x^2-x} - 9 &=& 0\\\nonumber 3^{x^2-x} &=& 3^2 \\\nonumber x^2-x -2 &=& 0 \end{eqnarray} \begin{eqnarray} x_{1,2} &=& \frac{1\pm \sqrt{1+8}}{2} = \frac{1\pm 3}{2}\\\nonumber x_1 &=& \frac{1+3}{2} = 2\\\nonumber x_2 &=& \frac{1-3}{2} = -1\end{eqnarray} - Example 16 - \(3^{x-1} + 2\cdot 3^{x} + 2\cdot 3^{x+1} = 9 \cdot 5^{x-1}\)
Solution - \begin{eqnarray}3^{x-1} + 2\cdot 3^{x} + 2\cdot 3^{x+1} &=& 9 \cdot 5^{x-1} \\\nonumber 3^x\left(\frac{1}{3} + 2+ 6 \right) &=& 3^2\cdot 5^{x-1}\\\nonumber 3^x\left(\frac{25}{3}\right) &=& 3^2\cdot 5^{x-1}\\\nonumber 5^2\cdot 3^x &=& 3^3\cdot 5^{x-1}/:(5^2\cdot 3^3)\\\nonumber3^{x-3} &=& 5^{x-3}\\\nonumber x-3 &=& 0 \\\nonumber x &=& 3\end{eqnarray} - Example 17 - Solve the equation \(4^x - 12\cdot 2^x + 32 = 0\) and multiply the obtained solutions.
Solution - \begin{eqnarray} 2^{2x} -12\cdot 2^x + 32 &=& 0 \Rightarrow 2^x = t \\\nonumber t^2 -12 t + 32 &=& 0\end{eqnarray} \begin{eqnarray}t_{1,2} &=& \frac{12 \pm \sqrt{144-128}}{2}\\\nonumber t_{1,2} &=& \frac{12 \pm 4}{2} \\\nonumber t_1 &=& \frac{12+4}{2} = 8\\\nonumber t_2 &=& \frac{12-4}{2} = 4\end{eqnarray} Now that the values of variable \(t\) are determined the values of variable \(x\) can be determined. \begin{eqnarray} 2^{x_1} &=& t_1 \Rightarrow 2^{x_1} = 8 \Rightarrow x_1 = 3\\\nonumber 2^{x_2} &=& t_2 \Rightarrow 2^{x_2} = 4 \Rightarrow x_2 = 2\end{eqnarray} The solutions of the equations are obtained and theri multiplication yields. $$x_1 \cdot x_2 = 3 \cdot 2 = 6$$ - Example 18 - Solve the equation \(7\cdot 2^\frac{2}{x^2}\cdot 3^\frac{2}{x^2} - 85 \cdot 6^{\frac{1}{x^2}-1} \cdot 7^\frac{1}{x^2} + 7^{\frac{2}{x^2}+1} = 0\), and calculte the sum and product of obtained solutions.
Solution - \begin{eqnarray}\end{eqnarray} - Example 19 - Solve the equation \(2^{x^2 -\frac{5}{7}x} = \sqrt[7]{4}\) and calculate the sum and the product of obtained solutions.
Solution - \begin{eqnarray}2^{x^2 -\frac{5}{7}x} &=& \sqrt[7]{4} \\\nonumber 2^{x^2 -\frac{5}{7}x} &=& 2^\frac{2}{7}\\\nonumber x^2-\frac{5}{7}x -\frac{2}{7} &=& 0\\\nonumber x_{1,2} &=& \frac{\frac{5}{7}\pm\sqrt{\frac{25}{49} + \frac{8}{7}}}{2} \\\nonumber x_{1,2} &=& \frac{\frac{5}{7}\pm \sqrt{\frac{25+56}{49}}}{2}\\\nonumber x_{1,2} &=& \frac{\frac{5}{7}\pm \sqrt{\frac{81}{49}}}{2}\\\nonumber x_{1,2} &=& \frac{\frac{5}{7}\pm \frac{9}{7}}{2}\\\nonumber x_1 &=& \frac{\frac{5}{7} + \frac{9}{7}}{2} = \frac{\frac{14}{7}}{2} = 1 \\\nonumber x_2 &=& \frac{\frac{5}{7}-\frac{9}{7}}{2} = \frac{-\frac{4}{7}}{2} = -\frac{2}{7}\end{eqnarray} The sum of obtained solutions $$x_1 + x_2 = 1 - \frac{2}{7} = \frac{7-2}{7} = \frac{5}{7} $$ The product of obtained solutions $$x_1 \cdot x_2 = 1\cdot \left(-\frac{2}{7}\right) = -\frac{2}{7} $$ - Example 20 - Solve the equation \(2^x\cdot 3^{x+1} \cdot 5^{x-3} = 21.6 \)
Solution - \begin{eqnarray}2^x\cdot 3^{x+1} \cdot 5^{x-3} &=& 21.6\\\nonumber 2^x\cdot 3^{x+1} \cdot 5^{x-3} &=& \frac{216}{10} \\\nonumber 2^x\cdot 3^{x+1} \cdot 5^{x-3} &=& \frac{108}{5}/\cdot 5 \\\nonumber 2^x\cdot 3^{x+1} \cdot 5^{x-2} &=& 2^2 3^3/:(2^2\cdot 3^3)\\\nonumber 2^{x-2}\cdot 3^{x-2}\cdot 5^{x-2} &=& 1\\\nonumber 30^{x-2} &=& 30^0 \\\nonumber x-2 &=& 0\\\nonumber x &=& 2\end{eqnarray} - Example 21 - Solve the equation \(\left(\frac{3}{2}\right)^x - \left(\frac{2}{3}\right)^{1-x} + \left(\frac{3}{2}\right)^{x-2} - \left(\frac{2}{3}\right)^{3-x} = 2\frac{7}{16}\)
Solution - \begin{eqnarray}\left(\frac{3}{2}\right)^x - \left(\frac{2}{3}\right)^{1-x} + \left(\frac{3}{2}\right)^{x-2} - \left(\frac{2}{3}\right)^{3-x} &=& 2\frac{7}{16} \\\nonumber \left(\frac{3}{2}\right)^x - \left(\frac{3}{2}\right)^{x-1} + \left(\frac{3}{2}\right)^{x-2} - \left(\frac{3}{2}\right)^{x-3} &=& \frac{39}{16}\\\nonumber \left(\frac{3}{2}\right)^x \left(1-\left(\frac{3}{2}\right)^{-1} + \left(\frac{3}{2}\right)^{-2} - \left(\frac{3}{2}\right)^{-3}\right) &=& \frac{39}{16}\\\nonumber \left(\frac{3}{2}\right)^x \left(\frac{27-9\cdot 2 +3\cdot 4 - 1\cdot 8}{27}\right) &=& \frac{39}{16}\\\nonumber \left(\frac{3}{2}\right)^x \left(\frac{13}{27}\right) &=& \frac{39}{16}/\cdot 27\\\nonumber \left(\frac{3}{2}\right)^x &=& \left(\frac{3}{2}\right)^4\\\nonumber x &=& 4\end{eqnarray} - Example 22 - Solve the equation \(2\sqrt{4^{|x+1|}} - 5\sqrt{6^|x+1|} + 3\sqrt{9^|x+1|} = 0\)
Solution - \begin{eqnarray}\end{eqnarray} - Example 23 -Solve the equation \(9\sqrt{3^{x^2 + |x|}} = \sqrt[3]{27^{-|x| +1 }}\)
Solution - \begin{eqnarray}\end{eqnarray} - Example 24 - Solve the equation \(2\cdot 3^|x| + 9^\frac{|x|}{2} - 27^\frac{|x|}{2} = 0\)
Solution - \begin{eqnarray}2\cdot 3^|x| + 9^\frac{|x|}{2} - 27^\frac{|x|}{2} &=& 0\\\nonumber 2\cdot 3^|x| + 3^|x| - 3^\frac{3|x|}{2} &=&= 0 \\\nonumber 3^{|x|+1} &=&3^\frac{3|x|}{2} \\\nonumber |x| + 1 &=& \frac{3|x|}{2}/\cdot 2 \\\nonumber 2|x| + 2 &=& 3|x|\\\nonumber -|x| &=& -2/\cdot (-1) \\\nonumber |x| &=& 2 \end{eqnarray} Since the variable x is under absolute value there are two equations i.e. two possible solutions. \begin{eqnarray} x_1 &=& 2 \\\nonumber x_2 &=& -2 \end{eqnarray} The product of the two solutions is equal to \(x_1\cdot x_2 = 2 \cdot(-2) = -4\) - Example 25 - Solve the equation \(4^{x^2-x + 1} - 8^x = 0\)
Solution - \begin{eqnarray}4^{x^2-x + 1} - 8^x &=& 0\\\nonumber 2^{2x^2 -2x + 2} &=& 2^{3x} \\\nonumber 2x^2 -2x +2 &=& 3x \\\nonumber 2x^2 -5x + 2 &=& 0 \end{eqnarray} \begin{eqnarray}x_{1,2} &=& \frac{5 \pm \sqrt{25-4\cdot 2\cdot 2}}{4} = \frac{5\pm 3}{4}\\\nonumber x_1 &=& \frac{5 +3 }{4} = \frac{8}{4} = 2\\\nonumber x_2 &=& \frac{5-3}{4} = \frac{5-3}{4} = \frac{1}{2}\end{eqnarray} - Example 26 - Solve the equation \(27^x = 9^{x+2}\)
Solution - \begin{eqnarray}27^x &=& 9^{x+2}\\\nonumber 3^{3x} &=& 3^{2x+4}\\\nonumber 3x &=& 2x + 4 \\\nonumber x &=& 4\end{eqnarray} - Example 27 - Solve the equation \(16^{x+4} = 32^{2x-10}\)
Solution - \begin{eqnarray}16^{x+4} &=& 32^{2x-10}\\\nonumber 2^{4x+16} &=& 2^{10x - 50} \\\nonumber 4x+16 &=& 10x - 50 \\\nonumber -6x &=& -66/:(-6)\\\nonumber x &=& 11\end{eqnarray} - Example 28 - Solve the equation \(12\left(\frac{4}{3}\right)^{-2x+1} - 25\left(\frac{3}{4}\right)^{x+1} + 9 = 0\), and calculate the sum of obtained solutions.
Solution - \begin{eqnarray}12\left(\frac{4}{3}\right)^{-2x+1} - 25\left(\frac{3}{4}\right)^{x+1} + 9 &=& 0\\\nonumber 9\left(\frac{3}{4}\right)^{2x} - 5^2 \left(\frac{3}{4}\right)^{x+1} + 3^2 &=& 0 \Rightarrow \left(\frac{3}{4}\right)^x = t \\\nonumber 9\cdot t^2 - 5^2\cdot \frac{3}{4} t + 9 &=& 0/\cdot 4 \\\nonumber 36t^2 - 75t + 36 &=& 0 \\\nonumber \end{eqnarray}\begin{eqnarray}t_{1,2} &=& \frac{75 \pm \sqrt{75^2 - 4\cdot 36^2}}{72} = \frac{75 \pm 21}{72} \\\nonumber t_1 &=& \frac{75 +21}{72} = \frac{96}{72} = \frac{4}{3}\\\nonumber t_2 &=& \frac{75-21}{72} = \frac{54}{72} = \frac{3}{4}\end{eqnarray}\begin{eqnarray}\left(\frac{3}{4}\right)^{x_1} &=& \frac{4}{3} \Rightarrow \left(\frac{3}{4}\right)^{x_1} &=& \left(\frac{3}{4}\right)^{-1}\Rightarrow x_1 = -1 \\\nonumber \left(\frac{3}{4}\right)^{x_2} &=& \left(\frac{3}{4}\right)^1 \Rightarrow x_2 = 1\end{eqnarray} Sum of the obtained solutions is equal to 0 (\(x_1 + x_2 = -1 +1 = 0 \).) - Example 29 -
Solution - \begin{eqnarray}\end{eqnarray} - Example 30 -
Solution - \begin{eqnarray}\end{eqnarray} - Example 31 - Solve the equation \(4^{2x^2+2x} = 8\)
Solution - \begin{eqnarray}4^{2x^2+2x} &=& 8\\\nonumber 2^{4x^2 + 4x} &=& 2^3 \\\nonumber 4x^2 + 4x - 3 &=& 0 \end{eqnarray} \begin{eqnarray}x_{1,2} &=& \frac{-4 \pm \sqrt{16+ 4\cdot 3 \cdot 4}}{8} \\\nonumber x_{1,2} &=& \frac{-4 \pm 8}{8} \\\nonumber x_1 &=& \frac{1}{2}\\\nonumber x_2 &=& -\frac{3}{2}\end{eqnarray} - Example 32 - Solve the equation \(-2^x = -4\)
Solution - \begin{eqnarray}-2^x &=& -4/\cdot (-1) \\\nonumber 2^x &=& 2^2 \\\nonumber x &=& 2\end{eqnarray} - Example 33 - Solve the equation \(4^{5-9x} = \frac{1}{8^{x-2}}\)
Solution - \begin{eqnarray}4^{5-9x} &=& \frac{1}{8^{x-2}} \\\nonumber 2^{10-18x} &=& 2^{-3x + 6} \\\nonumber 10-18x &=& -3x +6 \\\nonumber -15x &=& -4/:(-15) \\\nonumber x &=& \frac{4}{15}\end{eqnarray} - Example 34 - Solve the equation \(2^x = 2\sqrt{2}\)
Solution - \begin{eqnarray}2^x &=& 2\sqrt{2} \\\nonumber 2^x &=& 2^\frac{3}{2} \\\nonumber x &=& \frac{3}{2}\end{eqnarray} - Example 35 - Solve the equation \(10^x = 0.1 \cdot (1000)^{x-1}\)
Solution - \begin{eqnarray}10^x &=& 0.1 \cdot (1000)^{x-1}\\\nonumber 10^x &=& 10^{-1}\cdot 10^{3x-3}\\\nonumber 10^x &=& 10^{3x-4} \\\nonumber x-3x &=& -4 \\\nonumber -2x = -4/:(-2) \\\nonumber x = 2\end{eqnarray} - Example 36 - Solve the equation \(\frac{10^\frac{1}{3}\cdot 1000}{10^x} = 0.01\)
Solution - \begin{eqnarray}\frac{10^\frac{1}{3}\cdot 1000}{10^x} &=& 0.01\\\nonumber 10^\frac{1}{3}\cdot 10^3 \cdot 10^{-x} &=& \frac{1}{100}\\\nonumber 10^{\frac{10}{3} - x} &=& 10^{-2}\\\nonumber \frac{10}{3} - x &=& -2/\cdot 3 \\\nonumber 10-3x &=& -6 \\\nonumber -3x &=& -16 /:(-3) \\\nonumber x &=& \frac{16}{3}\end{eqnarray} - Example 37 - Solve the equation \(\sqrt[2x+4]{4^{x+8}} = \sqrt[6]{128}\)
Solution - \begin{eqnarray}\sqrt[2x+4]{4^{x+8}} &=& \sqrt[6]{128}\\\nonumber 2^{\frac{2x+16}{2x+4}} &=& 2^\frac{7}{6}\\\nonumber \frac{2x+16}{2x+4} &=& \frac{7}{6}\\\nonumber 6(2x+16) &=& 7(2x+4)\\\nonumber 12x + 96 &=& 14x +28 \\\nonumber -2x &=& -98 \\\nonumber x &=& 34\end{eqnarray} - Example 38 - Solve the equation \(2^x \cdot 5^x = 0.1\cdot\left(10^{x-1}\right)^5\)
Solution - \begin{eqnarray}2^x \cdot 5^x &=& 0.1\cdot\left(10^{x-1}\right)^5\\\nonumber 10^x &=& 10^{-1}\cdot 10^{5x-5}\\\nonumber 10^x &=& 10^{5x-6}\\\nonumber x &=& 5x -6 \\\nonumber -4x &=& -6 \\\nonumber x &=& \frac{3}{2}\end{eqnarray} - Example 39 - Solve the equation \(2^{-4+\sqrt{5x+1}} = 2^{2+\sqrt{5x+1}}\)
Solution - \begin{eqnarray}2^{-4+2\sqrt{5x+1}} &=& 2^{2+\sqrt{5x+1}}\\\nonumber -4 + 2\sqrt{5x+1} &=& 2+\sqrt{5x+1}\\\nonumber \sqrt{5x+1} &=& 6 /^2 \\\nonumber 5x + 1 &=& 36 \\\nonumber 5x &=& 35/:5 \\\nonumber x &=& 7\end{eqnarray} - Example 40 - Solve the equation \(9^{x+2} + 5\cdot 9^{x+1} = 14\)
Solution - \begin{eqnarray}9^{x+2} + 5\cdot 9^{x+1} &=& 14\\\nonumber 3^{2x+4} + 5 \cdot 3^{2x+2} &=& 7\cdot 2\\\nonumber 3^{2x}\left(3^4 + 5\cdot 3^2\right) = 7\cdot 2 \\\nonumber 3^{2x}(81+45) &=& 14 \\\nonumber 3^{2x}\cdot 126 = 14 /:126 \\\nonumber 3^{2x} &=& \frac{1}{9}\\\nonumber 2x &=& -2/:2 \\\nonumber x &=& -1\end{eqnarray} - Example 41 - Solve the equation \(\sqrt{5^{3x} + 19} - \sqrt{5^{3x} - 4} = 1\)
Solution - \begin{eqnarray}\sqrt{5^{3x} + 19} - \sqrt{5^{3x} - 4} &=& 1\\\nonumber \sqrt{5^{3x} + 19} &=& 1 +\sqrt{5^{3x} - 4} /^2 \\\nonumber 5^{3x} + 19 &=& 1 + 2\sqrt{5^{3x} - 4} + 5^{3x} - 4 \\\nonumber 5^{3x}(1-1) &=& 1 + 2\sqrt{5^{3x} -4}-23\\\nonumber 2\sqrt{5^{3x} - 4} &=& 22/:2 \\\nonumber \sqrt{5^{3x} - 4} &=& 11/^2\\\nonumber 5^{3x} - 4 &=& 121 \\\nonumber 5^{3x} &=& 125\\\nonumber 3x &=& 3/:3 \\\nonumber x&=& 1\end{eqnarray}
How to solve exponential equation ?
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment