Question

how would i solve 3^(x+4)=7^(x-1)

Answer 1

logarithms

Answer 2

\[3^{x+4}=7^{x-1}\]

Expand both sides the rule \(a^{b + c} = a^ba^c\)

Express the equation as a proportion in the following manner:
\(\dfrac{3^x}{7^x} = \dfrac{1}{7(3^4)}\)

Apply \(\dfrac{a^c}{b^c} = \left(\dfrac{a}{b}\right)^c\) to the left side:

\(\left(\dfrac{3}{7}\right)^x = \dfrac{1}{7(81)}\)

\(\left(\dfrac{3}{7}\right)^x = \dfrac{1}{567}\)

Now take the logs of both sides:

\(x \log(3/7) = \log (1/567)\)

\(x = \dfrac{\log(1/567)}{\log(3/7)}\)

Answer 3

Note that \(x\) can be simplified further to get

\(x = \dfrac{\log(7) + \log(3^4)}{\log(7) - \log(3)}\)