How do you solve for x in $2 {log}_{3} x = {log}_{3} 32 + {log}_{3} 2$ $2log_3x=log_3 32+log_3 2$ ?

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Answer 1 · 2017-02-09T19:24:49

$8 \sqrt{2}$ $8sqrt2$

For logarithms we have the following definition

${log}_{a} b = c \Rightarrow a^{c} = b - - - (1)$ $log_ab=c=>a^c=b---(1)$

we also have the standard results

${log}_{a} X + {log}_{a} Y = {log}_{a} X Y$ $log_aX+log_aY=log_aXY$ $- - - - (2)$ $----(2)$

${log}_{a} X - {log}_{a} Y = {log}_{a} (\frac{X}{Y}) - - - (3)$ $log_aX-log_aY=log_a(X/Y)---(3)$

${log}_{a} X^{n} = n {log}_{a} X$ $log_aX^n=nlog_aX$ color(white)(****) $- - - (4)$ $---(4)$

so we have

$2 {log}_{3} x = {log}_{3} 32 + {log}_{3} 8$ $2log_3x=log_(3)32+log_(3)8$

using $(2) on R H S$ $" "(2)" on "RHS$

$2 {log}_{3} x = {log}_{3} (32 \times 4)$ $2log_3x=log_(3)(32xx4)$

$2 {log}_{3} x = {log}_{3} (128)$ $2log_3x=log_(3)(128)$

$\Rightarrow {log}_{3} x^{2} = {log}_{3} 128$ $=>log_3x^2=log_(3)128$

$using (4) on L H S$ $"using "(4)" on "LHS$

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

$x^2=128$

$=>x=sqrt128$

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

How do you solve for x in 2log_3x=log_3 32+log_3 22log3x=log332+log322log_3x=log_3 32+log_3 2?