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log-2-x-log-4-x-log-16-x-7-




Question Number 32960 by mondodotto@gmail.com last updated on 08/Apr/18
 log_2 x+log_4 x+log_(16) x=7
$$\:\boldsymbol{\mathrm{log}}_{\mathrm{2}} \boldsymbol{{x}}+\boldsymbol{\mathrm{log}}_{\mathrm{4}} \boldsymbol{{x}}+\boldsymbol{\mathrm{log}}_{\mathrm{16}} \boldsymbol{{x}}=\mathrm{7} \\ $$
Answered by Joel578 last updated on 08/Apr/18
log_a^n   b^m  = (m/n) log_a  b          log_2  x + log_2^2   x + log_2^4   x = 7  log_(2 ) x + (1/2) log_2  x + (1/4) log_2  x = 7  (7/4) log_2  x = 7         log_2  x = 4  x = 2^4  = 16
$$\mathrm{log}_{{a}^{{n}} } \:{b}^{{m}} \:=\:\frac{{m}}{{n}}\:\mathrm{log}_{{a}} \:{b} \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{log}_{\mathrm{2}} \:{x}\:+\:\mathrm{log}_{\mathrm{2}^{\mathrm{2}} } \:{x}\:+\:\mathrm{log}_{\mathrm{2}^{\mathrm{4}} } \:{x}\:=\:\mathrm{7} \\ $$$$\mathrm{log}_{\mathrm{2}\:} {x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{log}_{\mathrm{2}} \:{x}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{log}_{\mathrm{2}} \:{x}\:=\:\mathrm{7} \\ $$$$\frac{\mathrm{7}}{\mathrm{4}}\:\mathrm{log}_{\mathrm{2}} \:{x}\:=\:\mathrm{7}\: \\ $$$$\:\:\:\:\:\:\mathrm{log}_{\mathrm{2}} \:{x}\:=\:\mathrm{4} \\ $$$${x}\:=\:\mathrm{2}^{\mathrm{4}} \:=\:\mathrm{16} \\ $$

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