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Hi Bunuel, why isn't (x-1)^2 is not treated like (a-b)^2 formula?

Thanks

Actually parenthesis were missing there. Edited, it should read: \((4^{(x-1)})^2=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2}\).

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink]
26 Jul 2015, 22:08

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