Solution for \log_7x+\log_{\frac 1 7}x^2=\log_{49}x-3

Maurice Maldonado

Maurice Maldonado

Answered question

2022-04-27

Solution for log7x+log17x2=log49x3

Answer & Explanation

otposlati9u8

otposlati9u8

Beginner2022-04-28Added 19 answers

What is the right solution for log7x+log17x2=log49x3
What logarithm identities used?
amisayq6t

amisayq6t

Beginner2022-04-29Added 18 answers

You will need the following identities: (For all a,b,c>0,n real)
(1) logab=xax=b
(2) logab=logcblogca
(3) nlogab=loga(bn)
Hence log17x2=logx2log17=logx2-log7=-log7x2=-2log7x
and log49x=log72x=12log7x
So log7x+log17x2=-log7x. Hence you have:
log7x=12log7x332log7x=3log7x=2

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