Simplify 2log(x+4)^2 - 2logx =1/4?

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Answer 1 · 2018-04-27T08:34:27

See below

The goal is to arrive to an expresion like $logA=logB$ from here and due to injectivity we will, have $A=B$

$2log(x+4)^2-2logx=1/4$

Using logarithm laws

$2(log(x+4)^2/x)=log10^(1/4)$

$log((x+4)^2/x)^2=log10^(1/4)$

Then $(x+4)^4/x^2=10^(1/4)$ sqrt in both sides

$(x+4)^2/x=10^(1/8)$ transposing terms we have

$x^2+(8-10^(1/8))x+16=0$ solving in x we will, have the solution