Solve

4^2x=2^5-x

and

9^x+1=27^2x-3

For the first one I keep getting like .4, and forthe life of me I can not figure out what im not doing right.

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- Dec 7th 2010, 09:54 AMabc10Exponential Equations
Solve

4^2x=2^5-x

and

9^x+1=27^2x-3

For the first one I keep getting like .4, and forthe life of me I can not figure out what im not doing right. - Dec 7th 2010, 09:57 AMe^(i*pi)
I assume you mean $\displaystyle 4^{2x} = 2^{5-x}$

Note that $\displaystyle 4^{2x} = 2^{4x}$

Now if two bases are equal then their exponents must also be equal so you get the linear equation $\displaystyle 4x = 5-x$ which is easy to solve

For question 2 use the fact that $\displaystyle 9 = 3^2 \text{ and } 27 = 3^3$ - Dec 7th 2010, 10:40 AMabc10
Thank you!