How do you solve for x in $2log_3x=log_3 32+log_3 2$ ?

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Answer 1 · 2017-02-09T19:24:49

$8sqrt2$

For logarithms we have the following definition

$log_ab=c=>a^c=b---(1)$

we also have the standard results

$log_aX+log_aY=log_aXY$ $----(2)$

$log_aX-log_aY=log_a(X/Y)---(3)$

$log_aX^n=nlog_aX$ color(white)(****) $---(4)$

so we have

$2log_3x=log_(3)32+log_(3)8$

using $" "(2)" on "RHS$

$2log_3x=log_(3)(32xx4)$

$2log_3x=log_(3)(128)$

$=>log_3x^2=log_(3)128$

$"using "(4)" on "LHS$

$because " the bases are the same we have:"$

$x^2=128$

$=>x=sqrt128$

$=>x=sqrt(64xx2)=8sqrt2$

How do you solve for x in 2log_3x=log_3 32+log_3 22log_3x=log_3 32+log_3 2?