How do you solve $3 {(2)}^{x - 2} + 1 = 100$ $3(2)^(x-2)+1=100$ ?

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How do you solve $3 {(2)}^{x - 2} + 1 = 100$ $3(2)^(x-2)+1=100$ ?

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Answer 1 · 2015-10-08T22:25:59

$x = {log}_{2} (33) + 2$ $x = log_2(33)+2$

Explanation:

As it is written, the answer does not look well rounded.
Here is the solution of the original equation
$3 \cdot 2^{x - 2} + 1 = 100$ $3*2^(x-2) + 1 = 100$

subtract $1$ $1$ from both sides:
$3 \cdot 2^{x - 2} = 99$ $3*2^(x-2) = 99$

divide by 3 both sides:
$2^{x - 2} = 33$ $2^(x-2) = 33$

apply ${log}_{2} ()$ $log_2()$ to both sides:
$x - 2 = {log}_{2} (33)$ $x-2 = log_2(33)$

add $2$ $2$ to both sides:
$x = {log}_{2} (33) + 2$ $x = log_2(33)+2$

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How do you solve $3 {(2)}^{x - 2} + 1 = 100$ $3(2)^(x-2)+1=100$ ?

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Explanation:

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How do you solve $3 {(2)}^{x - 2} + 1 = 100$ $3(2)^(x-2)+1=100$ ?