How do you solve for x in ${log}_{2} x = {log}_{4} (25)$ $log_2x=log_4(25)$ ?

Question

1993 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-01T20:48:05

$x = 5$ $x=5$

Using the formal definiton of a logarithm, we can take the right hand side to be:

${log}_{4} 25 = a \Rightarrow 4^{a} = 25$ $log_4 25=a=>4^a = 25$

Since $4$ $4$ is simply $2^{2}$ $2^2$ , we have:

$2^{2 a} = 25$ $2^(2a)=25$

Remember; $a$ $a$ is equal to ${log}_{4} 25$ $log_4 25$ , which is in turn equal to ${log}_{2} x$ $log_2 x$ .

$2^{2 {log}_{2} x} = 25$ $2^(2log_2x)=25$

Taking the binary logarithm of both sides:

$2 {log}_{2} x = {log}_{2} 25$ $2log_2x=log_2 25$

$\Rightarrow {log}_{2} x = \frac{1}{2} {log}_{2} 25$ $=> log_2x = 1/2log_2 25$

Knowing that $a {log}_{b} c = {log}_{b} c^{a}$ $alog_b c = log_b c^a$ , we reach the result we wished to get:

${log}_{2} x = {log}_{2} {(25)}^{\frac{1}{2}} = {log}_{2} 5 \Rightarrow x = 5$ $log_2x = log_2 (25)^(1/2) = log_2 5 => color(red)(x=5)$

How do you solve for x in log_2x=log_4(25)log2x=log4(25)log_2x=log_4(25)?