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

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Answer 1 · 2016-05-06T19:28:20

$x = 2+log_3(5)~~3.465$

$5-3^x = -40$

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.

$log_3(3^x) = log_3(45)$

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

$x = log_3(45)$

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

$x = log_3(9*5) = log_3(9)+log_3(5)$

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

$x = 2+log_3(5)~~3.465$

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