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

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Answer 1 · 2015-05-31T20:10:29

$4^{x} - 1 = 4^{- 1} \cdot 4^{x} = \frac{4^{x}}{4}$ $4^x-1 = 4^-1*4^x = 4^x/4$

$5^{x + 1} = 5^{1} \cdot 5^{x} = 5 \cdot 5^{x}$ $5^(x+1) = 5^1*5^x = 5*5^x$

Given

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

We have

$\frac{4^{x}}{4} = 5 \cdot 5^{x}$ $4^x/4 = 5*5^x$

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

$20 = \frac{4^{x}}{5^{x}} = {(\frac{4}{5})}^{x}$ $20 = 4^x/5^x = (4/5)^x$

Now take logs of both sides to get

$x \cdot log (\frac{4}{5}) = log (20)$ $x*log(4/5) = log(20)$

Divide both sides by $log (\frac{4}{5})$ $log(4/5)$ to get

$x = \frac{log (20)}{log (\frac{4}{5})} = \frac{log (4) + log (5)}{log (4) - log (5)}$ $x = log(20)/log(4/5) = (log(4)+log(5))/(log(4)-log(5))$

$log (2) ≅ 0.30103$ $log(2) ~= 0.30103$

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

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

$x ≅ \frac{0.60206 + 0.69897}{0.60206 - 0.69897}$ $x ~= (0.60206+0.69897)/(0.60206-0.69897)$

$= \frac{1.30103}{- 0.09691}$ $=1.30103/-0.09691$

$≅ - 13.425$ $~=-13.425$

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