Find-the-value-of-x-log-8-x-log-4-x-log-2-x-11-

Question Number 70008 by Shamim last updated on 30/Sep/19

Find the value of x .

\log_{8} x + \log_{4} x + \log_{2} x = 11 .

Answered by Maclaurin Stickker last updated on 30/Sep/19

{l o g}_{2^{3}} x + {l o g}_{2^{2}} x + {l o g}_{2} x = 11

\frac{1}{3} {l o g}_{2} x + \frac{1}{2} {l o g}_{2} x + {l o g}_{2} x = 11

(\frac{1}{3} + \frac{1}{2} + 1) {l o g}_{2} x = 11

\frac{11}{6} {l o g}_{2} x = 11

{l o g}_{2} x = 6

2^{6} = x

x = 64

Commented by Shamim last updated on 30/Sep/19

tnks a lot .

Commented by Shamim last updated on 30/Sep/19

plz explain with details -

\log_{2^{3}} x = \frac{1}{3} \log_{2} x

which law use for this ? ?

Commented by Tony Lin last updated on 30/Sep/19

{l o g}_{2^{3}} x = y

2^{3 y} = x

{l o g}_{2} x = 3 y

\frac{1}{3} {l o g}_{2} x = y = {l o g}_{2^{3}} x