How do you solve $3 {log}_{5} x - {log}_{5} 4 = {log}_{5} 16$ $3Log_5 x - Log_5 4 = Log_5 16$ ?

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Answer 1 · 2016-06-15T07:18:10

$x = 4$ $x=4$

$3 {log}_{5} x - {log}_{5} 4 = {log}_{5} 16$ $3log_5x-log_(5)4=log_(5)16$

or ${log}_{5} x^{3} - {log}_{5} 4 = {log}_{5} 16$ $log_5x^3-log_(5)4=log_(5)16$

or $\frac{{log}_{5} x^{3}}{4} = {log}_{5} 16$ $log_5x^3/4=log_(5)16$

or $\frac{x^{3}}{4} = 16$ $x^3/4=16$

or $x^{3} = 64$ $x^3=64$

or $x = 4$ $x=4$

How do you solve 3Log_5 x - Log_5 4 = Log_5 163log5x−log54=log5163Log_5 x - Log_5 4 = Log_5 16?