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

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Answer 1 · 2016-05-06T19:28:20

$x = 2 + {log}_{3} (5) \approx 3.465$ $x = 2+log_3(5)~~3.465$

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

Subtract $5$ $5$ from each side of the equation.

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

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

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

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

${log}_{3} (3^{x}) = {log}_{3} (45)$ $log_3(3^x) = log_3(45)$

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

$x = {log}_{3} (45)$ $x = log_3(45)$

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

$x = {log}_{3} (9 \cdot 5) = {log}_{3} (9) + {log}_{3} (5)$ $x = log_3(9*5) = log_3(9)+log_3(5)$

Apply the rule ${log}_{a} (a^{x}) = x$ $log_a(a^x) = x$ to ${log}_{3} (9) = {log}_{3} (3^{2})$ $log_3(9)=log_3(3^2)$

$x = 2 + {log}_{3} (5) \approx 3.465$ $x = 2+log_3(5)~~3.465$

How do you solve 5−3x=−405 - 3^x = - 40?