How to solve for f?

Question

The question asks to solve for the variable: $$2=6(3^{4f-2})$$

I am not quite sure how to solve for $f$ because the bases on either side cannot be made equal. Here is an example of a similar equation that I was able to understand: $$2(4^{v+1})=1$$ $$2[(2^{2})^{v+1}]=2^{0}$$ $$2(2^{2v+3})=2^{0}$$

$$\therefore2v+3=0$$ $$2v=-3$$ $$2v=-3\over2$$ $$v=-3\over2$$

Answer 1

It is simple, it can be written as \begin{equation} 3^{-1} = 3^{4f-2} \end{equation} and hence $4f-2=-1$

Answer 2

You do not need the bases on each side to be equal to solve this problem. First let's clean up the RHS as much as possible. $$2=6(3^{4f-2}) = 6(3^{4f}3^{-2}) = 6((3^{f})^4 3^{-2})$$ The properties I used were: $a^{xy} = (a^{x})^{y}$ and $a^{b+c} = a^ba^c$ for a base $a$. Dividing both sides of our equation by $6$ yields $$\begin{align} \implies \frac{1}{3} =(3^{f})^4 3^{-2} \\ \implies 3 = (3^{f})^4 \\ \implies \sqrt[4]{3} = 3^f \\ \implies \log_3\left(\sqrt[4]{3}\right)=f \end{align}$$ Now can you apply logarithm properties to simplify $\log_3\left(\sqrt[4]{3}\right)$? Things will cancel quite nicely.

Answer 3

you can factor the $6$ on the right hand side and rewrite your equations as $$2 = 2 \times 3 \times 3^{4f-2} \to 1 = 3^{4f-1}\to 4f = 1\to f = \frac 14. $$