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2log base 4x-log base 4(x+5)=2
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\[2log_4(x)-log_4(x+5) = 2?\]
Yes
\[2log_4(x)-log_4(x+5) = log_4(x^2) - log_4(x+5) = \]
do you get this step?
Yes thx
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\[log(a)-log(b) = log(\frac{a}{b})\]
so \[log(x^2) - log(x+5)= log(\frac{x^2}{x+5})\] so \[log_4(\frac{x^2}{x+5})=2\] this means \[4^2 = \frac{x^2}{x+5}\] so \[16(x+5) = x^2 \rightarrow x^2-16x-80=0\] can you solve \[x^2-16x-80=0?\]
(x-20)(x+4)=0 x=20 and x=-4 but \[log_4(x) \] is not defined for x<= 0 so, we take x=20 to be the solution
Thanks!
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