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Solve this equation algebraically.

Solve this equation algebraically.-example-1

1 Answer

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Write both sides as powers of 16:


\log_4(2-x)=\log_(16)(13-4x) \implies 16^(\log_4(2-x))=16^{\log_(16)(13-4x)}

Since 16 = 4², we can rearrange the left side to get


16^(\log_4(2-x))=(4^2)^(\log_4(2-x))=4^(2\log_4(2-x))=4^(\log_4(2-x)^2)

Then, recalling that
b^(\log_ba)=a, we have


4^(\log_4(2-x)^2)=16^{\log_(16)(13-4x)}\implies(2-x)^2=13-4x

Now solve for x :


(2-x)^2=13-4x


4-4x+x^2=13-4x


x^2-9=0


(x+3)(x-3)=0


\implies x=-3\text{ or }x=3

Notice that if x = 3, then log₄(2 - x) = log₄(-1) is undefined, so we throw out this solution. We don't run into this problem with x = -3, so that's the only (real) solution.

User Oleg Ushakov
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