How do you solve $4^(x-1)=5^(x+1)$ ?

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Answer 1 · 2015-05-31T20:10:29

$4^x-1 = 4^-1*4^x = 4^x/4$

$5^(x+1) = 5^1*5^x = 5*5^x$

Given

$4^(x-1) = 5^(x+1)$

We have

$4^x/4 = 5*5^x$

Multiply both sides by $4$ and divide both sides by $5^x$ to get

$20 = 4^x/5^x = (4/5)^x$

Now take logs of both sides to get

$x*log(4/5) = log(20)$

Divide both sides by $log(4/5)$ to get

$x = log(20)/log(4/5) = (log(4)+log(5))/(log(4)-log(5))$

$log(2) ~= 0.30103$

$log(4) = 2log(2) ~= 0.60206$

$log(5) = 1 - log(2) ~= 0.69897$

$x ~= (0.60206+0.69897)/(0.60206-0.69897)$

$=1.30103/-0.09691$

$~=-13.425$

How do you solve 4^(x-1)=5^(x+1) 4^(x-1)=5^(x+1)?