# algebra

3^(x+4)=2^(1-3x)

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1. Take logarithm on both sides and use the exponent rule of logarithms:
log(ab) = b*log(a)

So applying the rule to the given equation:
3^(x+4)=2^(1-3x)
log(3^(x+4)) = log(2^(1-3x))
(x+4)log(3) = (1-3x)log(2)
Solve the resulting linear equation after substituting numerical values of log(3) and log(2). You can use either natural log (to the base e) or common log (to the base 10).
I get x=-1.2 approximately.

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2. i'm sorry i don't understand the last part :(

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3. Evaluate log(3) and log(2) using logarithm to the base 10.
log103 = 0.4771
log102 = 0.3010

(x+4)log(3) = (1-3x)log(2)
0.4771(x+4) = 0.3010(1-3x)

Solving for x, I get x=1.16 approx.

If you had used logarithm to the base e (natural log), you would have got the same answer of x=1.16 approx.

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