How do you solve $2^{x} = 20$ $2^x = 20$ ?

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Answer 1 · 2016-02-29T23:16:08

Use basic properties of logarithms to remove $x$ $x$ from the exponent to find that

$x = {log}_{2} (20) \approx 4.322$ $x = log_2(20)~~4.322$

We will use the following properties of logarithms:

With these, we have

$2^{x} = 20$ $2^x = 20$

$\Rightarrow {log}_{2} (2^{x}) = {log}_{2} (20)$ $=> log_2(2^x) = log_2(20)$

$\Rightarrow x {log}_{2} (2) = {log}_{2} (20)$ $=>xlog_2(2)=log_2(20)$

$\Rightarrow x (1) = {log}_{2} (20)$ $=>x(1)=log_2(20)$

$:.x = log_2(20)~~4.322$

How do you solve 2^x = 202x=20 2^x = 20?