How would I solve this? 2^2x - 2^(x+1) = 0 Having trouble with this too 2^2x - 12 * 2^x + 32 = 0
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$\displaystyle 2^{2x}=2^{x+1}$ $\displaystyle 2x=x+1$ $\displaystyle x=1$
For second one $\displaystyle (2^x - 8)(2^x - 4) = 0$ $\displaystyle 2^x =8$ gives $\displaystyle x=3$ $\displaystyle 2^x =4$ gives $\displaystyle x=2$
Thanks Glaysher!
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