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Gelneren [198K]
3 years ago
14

I NEED HELP!! please show all work

Mathematics
1 answer:
PtichkaEL [24]3 years ago
5 0

Take the logarithm of both sides. The base of the logarithm doesn't matter.

4^{5x} = 3^{x-2}

\implies \log 4^{5x} = \log 3^{x-2}

Drop the exponents:

\implies 5x \log 4 = (x-2) \log 3

Expand the right side:

\implies 5x \log 4 = x \log 3 - 2 \log 3

Move the terms containing <em>x</em> to the left side and factor out <em>x</em> :

\implies 5x \log 4 - x \log 3 = - 2 \log 3

\implies x (5 \log 4 - \log 3) = - 2 \log 3

Solve for <em>x</em> by dividing boths ides by 5 log(4) - log(3) :

\implies \boxed{x = -\dfrac{ 2 \log 3 }{ 5 \log 4 - \log 3 }}

You can stop there, or continue simplifying the solution by using properties of logarithms:

\implies x = -\dfrac{ \log 3^2 }{ \log 4^5 - \log 3 }

\implies x = -\dfrac{ \log 9 }{ \log 1024 - \log 3 }

\implies \boxed{x = -\dfrac{ \log 9 }{ \log \frac{1024}3 }}

You can condense the solution further using the change-of-base identity,

\implies \boxed{x = -\log_{\frac{1024}3}9}

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------------------------
What I did to solve it was....

1. Divide top fraction by bottom fraction to get decimal
1 ÷ 2 = 0.5

2. Then multiply 0.5 by its self (the same as putting it to the ^2)
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