tylerjcarr tylerjcarr

22-07-2018
Mathematics

contestada

This was my very last homework question and i'm stumped.
log_4(x+1)-log_4(x)=3

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konrad509 konrad509

22-07-2018

[tex] \log_4(x+1)-\log_4x=3 \qquad(x>0)\\
\log_4\dfrac{x+1}{x}=3\\
\dfrac{x+1}{x}=4^3\\
\dfrac{x+1}{x}=64\\
x+1=64x\\
63x=1\\
x=\dfrac{1}{63} [/tex]

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