How do you condense ${log}_{6} (x + 4) + \frac{1}{2} {log}_{6} x$ $log_(6)(x+4)+1/2log_(6)x$ ?

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Answer 1 · 2016-08-04T06:04:26

${log}_{6} ((x + 4) \sqrt{x})$ $log_6((x+4)sqrtx)$

Since $n {log}_{b} a = {log}_{b} a^{n}$ $n log_b a=log_b a^n$ , you have

$\frac{1}{2} {log}_{6} x = {log}_{6} x^{\frac{1}{2}} = {log}_{6} \sqrt{x}$ $1/2log_6 x=log_6 x^(1/2)=log_6sqrt(x)$

Since ${log}_{b} a + {log}_{b} c = {log}_{b} (a c)$ $log_b a+log_b c=log_b (ac)$ , you have

${log}_{6} (x + 4) + {log}_{6} \sqrt{x} = {log}_{6} ((x + 4) \sqrt{x})$ $log_6(x+4)+log_6sqrtx=log_6((x+4)sqrtx)$

How do you condense log_(6)(x+4)+1/2log_(6)xlog6(x+4)+12log6xlog_(6)(x+4)+1/2log_(6)x?