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Log6^(4x-5)=Log6^(2x+1)

1 Answer

3 votes

Answer:


x = 3

Explanation:

Given


\log6^((4x-5)) =\log6^((2x+1))

Required

Solve for x

We have:


\log6^((4x-5)) =\log6^((2x+1))

Remove log6 from both sides


(4x-5) = (2x+1)

Collect like terms


4x - 2x = 5 + 1


2x = 6

Divide by 2


x = 3

User Artem Fedosov
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