How do you solve $8^(2x-5)=5^(x+1)$ ?

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Answer 1 · 2016-12-10T14:03:40

$x = (5ln(8)+ln(5))/(2ln(8)-ln(5))~~4.7095$

Using the property of logarithms that $log(a^x) = xlog(a)$ , we have

$8^(2x-5) = 5^(x+1)$

$=> ln(8^(2x-5)) = ln(5^(x+1))$

$=> (2x-5)ln(8) = (x+1)ln(5)$

$=> 2ln(8)x - 5ln(8) = ln(5)x+ln(5)$

$=>2ln(8)x - ln(5)x = 5ln(8)+ln(5)$

$=> (2ln(8)-ln(5))x = 5ln(8)+ln(5)$

$:. x = (5ln(8)+ln(5))/(2ln(8)-ln(5))~~4.7095$

How do you solve 8^(2x-5)=5^(x+1)8^(2x-5)=5^(x+1)?