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

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Answer 1 · 2016-12-10T14:03:40

$x = \frac{5 ln (8) + ln (5)}{2 ln (8) - ln (5)} \approx 4.7095$ $x = (5ln(8)+ln(5))/(2ln(8)-ln(5))~~4.7095$

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

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

$\Rightarrow ln (8^{2 x - 5}) = ln (5^{x + 1})$ $=> ln(8^(2x-5)) = ln(5^(x+1))$

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

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

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

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

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

How do you solve 8^(2x-5)=5^(x+1)82x−5=5x+18^(2x-5)=5^(x+1)?