How do I solve this fractional indices equation (3^(5x+2))/(9^(1-x))=(27^(4+3x))/(729)

Mackenzie Hawkins 2022-09-23 Answered
How do I solve this fractional indices equation 3 5 x + 2 9 1 x = 27 4 + 3 x 729 ?
Solve the equation 3 5 x + 2 9 1 x = 27 4 + 3 x 729
I thought that the best way of approaching this would be to rewrite everything using 3 as the base of the exponents, hence creating an equivalence which would allow me to equate numerators to numerators, and denominators to denominators. Doing this yields:
3 5 x + 2 3 2 ( 1 x ) = 3 3 ( 4 + 3 x ) 3 6
Equating the exponents of each numerator:
5 x + 2 = 3 ( 4 + 3 x )
5 x + 2 = 12 + 9 x
4 x = 10
x = 10 4 = 5 2
Doing this for the denominator yields a different value of x:
2 ( 1 x ) = 6
2 2 x = 6
2 x = 4
x = 2
Why is that I'm obtaining two different values of x? Further to this, the solution in the book states the answer as x = 3, what am I doing wrong?
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Answers (2)

Nancy Ewing
Answered 2022-09-24 Author has 7 answers
Recall that the property
a f ( x ) = a g ( x ) f ( x ) = g ( x )
holds because the exponential function is injective. Therefore, you always need to rewrite both sides of the equation as powers of a before you can equate the exponents.
In your situation, you can apply the property:
a m a n = a m n
and thus write
3 5 x + 2 2 ( 1 x ) = 3 3 ( 4 + 3 x ) 6 .
Then you can solve the equation
5 x + 2 2 ( 1 x ) = 3 ( 4 + 3 x ) 6.
Now, if you're still wondering why you can't just equate the exponents of the numerators and those of the denominators, let me show you a simple counterexample to explain why that might not work in general:
3 7 3 5 = 3 8 3 6
holds because if you compute the quotients they are both equal to 9, but 7 8 and 5 6.
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Seamus Mcknight
Answered 2022-09-25 Author has 3 answers
First subtract the exponents both side and then equate.
3 5 x + 2 2 ( 1 x ) = 3 3 ( 4 + 3 x ) 6
5 x + 2 2 ( 1 x ) = 3 ( 4 + 3 x ) 6 x = 3
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