Solution to Equations Involving Variable Exponent

$9^x - 6(9^{-x}) - 1 = 0$

$9^x - \dfrac{6}{9^x} - 1 = 0$
 

Multiply both sides by 9^x
$9^{2x} - 6 - 9^x = 0$

$9^{2x} - 9^x - 6 = 0$
 

Let z = 9^x
$z^2 - z - 6 = 0$

$(z - 3)(z + 2) = 0$

$z = 3 ~ \text{ and/or } ~ -2$
 

For z = 3

$9^x = 3$

$9^x = 9^{1/2}$

$x = \frac{1}{2}$
 

Check the given equation for x = 1/2
$9^{1/2} - 6(9^{-1/2}) - 1 = 0$

$3 - 2 - 1 = 0$   ←   okay!

For z = -2

$9^x = -2$

$\ln 9^x = \ln (-2)$

$x \ln 9 = \ln (-2)$

$x = \dfrac{\ln (-2)}{\ln 9}$

$x = \text{not a real number}$

 

Hence, x = 1/2           answer

$2^{x + 1} \cdot 3^x = 5^{x + 3}$

$\ln \left( 2^{x + 1} \cdot 3^x \right) = \ln 5^{x + 3}$

$\ln 2^{x + 1} + \ln 3^x = \ln 5^{x + 3}$

$(x + 1) \ln 2 + x \ln 3 = (x + 3) \ln 5$

$x \ln 2 + \ln 2 + x \ln 3 = x \ln 5 + 3 \ln 5$

$x \ln 2 + x \ln 3 - x \ln 5 = 3 \ln 5 - \ln 2$

$x \ln 2 + x \ln 3 - x \ln 5 = \ln 5^3 - \ln 2$

$x (\ln 2 + \ln 3 - \ln 5) = \ln 125 - \ln 2$

$x \ln \dfrac{2(3)}{5} = \ln \dfrac{125}{2}$

$x \ln 1.2 = \ln 62.5$

$x = \dfrac{\ln 62.5}{\ln 1.2}$

$x = 22.68$           answer