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4^9n+4=256
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\(\bf \large 4^{9n}+4=256 \quad ?\)
Nope. 4+^9n+4 (<-- All to the power of) The only numbers that aren't powers are the first "4" and the "256" at the end. Does that help?
\(\bf \large {4^{9n+4}=256\\ \quad \\ \textit{let us use the log cancellation rule of }\quad log_aa^x = x\\ \quad \\ 4^{9n+4}=256\implies log_4(4^{9n+4})=log_4(256)\\\quad \\\implies 9n+4=log_4(256)\\ \quad \\ 9n = log_4(256)-4\implies n = \cfrac{log_4(256)-4}{9}}\)
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