How do you solve ${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2x+log_4x=log_2 5$ ?

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How do you solve ${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2x+log_4x=log_2 5$ ?

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Answer 1 · 2016-04-09T06:40:03

${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2x+log_4x=log_2 5$
$\Rightarrow {log}_{2} x + \frac{1}{{log}_{x} 4} = {log}_{2} 5$ $=>log_2x+1/log_x4=log_2 5$
$\Rightarrow {log}_{2} x + \frac{1}{{log}_{x} 2^{2}} = {log}_{2} 5$ $=>log_2x+1/log_x2^2=log_2 5$
$\Rightarrow {log}_{2} x + \frac{1}{2 {log}_{x} 2} = {log}_{2} 5$ $=>log_2x+1/(2log_x2)=log_2 5$
$\Rightarrow {log}_{2} x + \frac{1}{2} \times {log}_{2} x = {log}_{2} 5$ $=>log_2x+1/2xxlog_2x=log_2 5$
$\Rightarrow {log}_{2} x + {log}_{2} x^{\frac{1}{2}} = {log}_{2} 5$ $=>log_2x+log_2x^(1/2)=log_2 5$
$\Rightarrow {log}_{2} (x \cdot x^{\frac{1}{2}}) = {log}_{2} 5$ $=>log_2(x*x^(1/2))=log_2 5$
$\Rightarrow (x \cdot x^{\frac{1}{2}}) = 5$ $=>(x*x^(1/2))=5$
$\Rightarrow (x^{\frac{3}{2}}) = 5$ $=>(x^(3/2))=5$
$:. x=5^(2/3)$

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How do you solve ${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2x+log_4x=log_2 5$ ?

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How do you solve ${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2x+log_4x=log_2 5$ ?