How do you solve ${log}_{2} x + {log}_{8} 32 x = 1$ $log_2 x+log_8 32x=1$ ?

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How do you solve ${log}_{2} x + {log}_{8} 32 x = 1$ $log_2 x+log_8 32x=1$ ?

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Answer 1 · 2016-04-23T09:13:11

${log}_{2} x + {log}_{8} 32 x = 1$ $log_2x+log_8 32x=1$
$\Rightarrow {log}_{2} x + {log}_{8} 32 + {log}_{8} x = 1$ $=>log_2x+log_8 32+log_8x=1$
$\Rightarrow {log}_{2} x + {log}_{8} 2^{5} + {log}_{8} x = 1$ $=>log_2x+log_8 2^5+log_8x=1$
$\Rightarrow {log}_{2} x + 5 {log}_{8} 2 + {log}_{8} x = 1$ $=>log_2x+5log_8 2+log_8x=1$
$\Rightarrow {log}_{2} x + 5 \times \frac{1}{{log}_{2} 8} + \frac{1}{{log}_{x} 8} = 1$ $=>log_2x+5xx1/log_2 8+1/log_x 8=1$
$\Rightarrow {log}_{2} x + 5 \times \frac{1}{{log}_{2} 2^{3}} + \frac{1}{{log}_{x} 2^{3}} = 1$ $=>log_2x+5xx1/log_2 2^3+1/log_x 2^3=1$
$\Rightarrow {log}_{2} x + 5 \times \frac{1}{3 {log}_{2} 2} + \frac{1}{3 {log}_{x} 2} = 1$ $=>log_2x+5xx1/(3log_2 2)+1/(3log_x 2)=1$
$\Rightarrow {log}_{2} x + \frac{5}{3} + \frac{1}{3} \times {log}_{2} x = 1$ $=>log_2x+5/3+1/3xxlog_2 x=1$
$\Rightarrow 3 {log}_{2} x + 5 + {log}_{2} x = 3$ $=>3log_2x+5+log_2 x=3$
$\Rightarrow 4 {log}_{2} x = 3 - 5 = - 2$ $=>4log_2x=3-5=-2$
$\Rightarrow {log}_{2} x = - \frac{1}{2}$ $=>log_2 x=-1/2$
$\Rightarrow x = 2^{- \frac{1}{2}} = \frac{1}{\sqrt{2}}$ $=>x=2^(-1/2)=1/sqrt2$

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How do you solve ${log}_{2} x + {log}_{8} 32 x = 1$ $log_2 x+log_8 32x=1$ ?

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How do you solve ${log}_{2} x + {log}_{8} 32 x = 1$ $log_2 x+log_8 32x=1$ ?