\[ If\ \ 3^k+3^k=(3^9)^{3^9}-3^k,\ \ \ then\ \ \ \ k=? \]

VJesus12 wrote:$$If\ \ 3^k+3^k=(3^9)^{3^9}-3^k,\ \ \ then\ \ \ \ k=?$$ (A) 11/3
(B) 11/2
(C) 242
(D) 3^10
(E) 3^11 - 1

The OA is the option E.

How can I rewrite the given expression to get the correct answer? I could solve the powers. <i class="em em-frowning"></i>

Hello Vjesus12.

You can rewrite the powers as follows: $$3^k+3^k=(3^9)^{3^9}-3^k\ \ \ \Rightarrow\ \ \ \ 2\cdot3^k=\left(3\right)^{9\cdot3^9}-3^k$$ $$\Rightarrow\ \ \ \ 2\cdot3^k+3^k=\left(3\right)^{3^2\cdot3^9}$$ $$\Rightarrow\ \ \ \ 3\cdot3^k=\left(3\right)^{3^{11}}$$ $$\Rightarrow\ \ \ \ 3^{k+1}=\left(3\right)^{3^{11}}$$ $$\Rightarrow\ \ k+1=3^{11}$$ $$\Rightarrow\ \ k=3^{11}-1.$$ That way we get that the correct answer is the option E .

I hope it helps you. <i class="em em-smiley"></i>