Question

Solve using logarithms 7^(x-2)=5^(3-x)

Answer 1

when you log something like say \[x^{5}\] and you log it to make \[\log x^{5} \]
you can pull the exponent down to the front like so
\[5\log x \]

Answer 2

if that is the case, you can pull the x terms in the exponents and solve.

Answer 3

pull the x terms in the exponents to the front*

Answer 4

but doesn't that kinda take you backwards since you have to solve for x?

Answer 5

but you can't exactly solve for x while having them in the exponents can you? if you pull the (x-2)to the front (x-2)log7 = (3-x)log5, you can isolate like terms that contain x and the rest would just be in decimal numbers

Answer 6

\(\large 7^{x-2} = 5^{3-x}\)
Take logs on both sides:
\(\large \log(7^{x-2}) = \log(5^{3-x})\)
\(\large (x-2)\log(7) = (3-x)\log(5)\)
\(\large x\log(7)-2\log(7) = 3\log(5)-x\log(5)\)
Add \(\large x\log(5)\) to both sides:
\(\large x\log(7) + x\log(5) -2\log(7) = 3\log(5)\)
Add \(\large 2\log(7)\) to both sides:
\(\large x\log(7) + x\log(5) = 3\log(5) + 2\log(7) \)
\(\large x(\log(7) + \log(5)) = 3\log(5) + 2\log(7) \)
\(\large x\log(35) = 3\log(5) + 2\log(7) \)
\(\large x = \Large \frac{3\log(5) + 2\log(7)}{\log(35)} \)