How do you solve $2^{x - 3} > 5$ $2^(x-3)>5$ ?

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How do you solve $2^{x - 3} > 5$ $2^(x-3)>5$ ?

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Answer 1 · 2016-12-07T11:27:12

I found: $x > 3 + {log}_{2} (5)$ $x>3+log_2(5)$

Explanation:

We can try taking the log of both sides (base $2$ $2$ ):
${log}_{2} (2^{x - 3}) > {log}_{2} (5)$ $log_2(2^(x-3))>log_2(5)$
applying the definition of log we get:
$x - 3 > {log}_{2} (5)$ $x-3>log_2(5)$
rearrange:
$x > 3 + {log}_{2} (5)$ $x>3+log_2(5)$
or:
$x > 5.3219$ $x>5.3219$

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How do you solve $2^{x - 3} > 5$ $2^(x-3)>5$ ?

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