How do you solve 5 - 3^x = - 40?

May 6, 2016

$x = 2 + {\log}_{3} \left(5\right) \approx 3.465$

Explanation:

$5 - {3}^{x} = - 40$

Subtract $5$ from each side of the equation.

$- {3}^{x} = - 45$

Multiply both sides of the equation by $- 1$.

${3}^{x} = 45$

Take the base-$3$ logarithm of each side of the equation.

${\log}_{3} \left({3}^{x}\right) = {\log}_{3} \left(45\right)$

Apply the rule ${\log}_{a} \left({a}^{x}\right) = x$ to the left hand side.

$x = {\log}_{3} \left(45\right)$

Apply the rule $\log \left(a b\right) = \log \left(a\right) + \log \left(b\right)$ to the right hand side.

$x = {\log}_{3} \left(9 \cdot 5\right) = {\log}_{3} \left(9\right) + {\log}_{3} \left(5\right)$

Apply the rule ${\log}_{a} \left({a}^{x}\right) = x$ to ${\log}_{3} \left(9\right) = {\log}_{3} \left({3}^{2}\right)$

$x = 2 + {\log}_{3} \left(5\right) \approx 3.465$