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