Exponent Question

Question

This is the equation I am trying to solve:

$$4^{27} + 4^{27} + 4^{27} + 4^{27} = 4^{y-1}.$$

I understand that I should convert to base $4$ and solve but I am not sure about how to do that, are there any other ways of solving this? Thanks

Answer 1

$$4^{27}+4^{27}+4^{27}+4^{27} = 4 \cdot 4^{27} = 4^{28}=4^{29-1}$$

Answer 2

Just note that this implies $4 \cdot 4^{27} = 4^{28} = 4^{y-1}$, so $y =29$.

Answer 3

$$4^{27}+4^{27}+4^{27}+4^{27} = 4^{27}(1+1+1+1) = 4^{27}\cdot4 =4^{28}=4^{29-1}$$

$$4^{y-1}=4^{29-1}$$

Thus $y=29$