How do you solve ${log}_{5} 6 + {log}_{5} (2 x^{2}) = {log}_{5} 48$ $log_5 6 +log_5 (2x^2)=log_5 48$ ?

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Answer 1 · 2016-10-05T14:15:36

$x = \pm 2$ $x=+-2$

Use ${log}_{b} m + {log}_{b} n = {log}_{b} (m n)$ $log_b m+log_b n = log_b(mn)$

Here, ${log}_{5} 6 + {log}_{5} (2 x^{2}) = {log}_{b} (12 x^{2}) = {log}_{5} 48$ $log_5 6+log_5(2x^2)= log_b (12x^2)=log_5 48$

Log function is a bijective (single valued) function. So,

$12 x^{2} = 48$ $12x^2=48$ , and so,

$x^{2} = 4 \to x = \pm 2$ $x^2=4 to x =+-2$

How do you solve log_5 6 +log_5 (2x^2)=log_5 48log56+log5(2x2)=log548log_5 6 +log_5 (2x^2)=log_5 48?