If $64^{x} - 1 = 16^{\frac{x}{2}}$ $64^x-1=16^(x/2)$ , find $x$ $x$ ?

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If $64^{x} - 1 = 16^{\frac{x}{2}}$ $64^x-1=16^(x/2)$ , find $x$ $x$ ?

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Answer 1 · 2017-11-09T06:55:20

$x = \frac{3}{2}$ $x=3/2$

Explanation:

It is assumed here that we are given $64^{x - 1} = 16^{\frac{x}{2}}$ $64^(x-1)=16^(x/2)$ and not $64^{x} - 1 = 16^{\frac{x}{2}}$ $64^x-1=16^(x/2)$

$64^{x - 1} = 16^{\frac{x}{2}}$ $64^(x-1)=16^(x/2)$

or ${ln 64}^{x - 1} = {ln 16}^{\frac{x}{2}}$ $ln64^(x-1)=ln16^(x/2)$

or $(x - 1) ln 64 = (\frac{x}{2}) ln 16$ $(x-1)ln64=(x/2)ln16$

or $(x - 1) (\frac{2}{x}) = \frac{ln 16}{ln 64}$ $(x-1)(2/x)=ln16/ln64$

or $(x - 1) \frac{1}{x} = \frac{ln 16}{2 ln 64}$ $(x-1)1/x=ln16/(2ln64)$

or $1 - \frac{1}{x} = \frac{{ln 2}^{4}}{2 {ln 2}^{6}}$ $1-1/x=ln2^4/(2ln2^6)$

or $- \frac{1}{x} = \frac{4 ln 2}{12 ln 2} - 1$ $-1/x=(4ln2)/(12ln2)-1$

or $- \frac{1}{x} = \frac{1}{3} - 1 = - \frac{2}{3}$ $-1/x=1/3-1=-2/3$

or $\frac{1}{x} = \frac{2}{3}$ $1/x=2/3$

i.e. $x = \frac{3}{2}$ $x=3/2$

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If $64^{x} - 1 = 16^{\frac{x}{2}}$ $64^x-1=16^(x/2)$ , find $x$ $x$ ?

1 Answer

Explanation:

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If 64x−1=16x264^x-1=16^(x/2), find xx?

1 Answer

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If $64^{x} - 1 = 16^{\frac{x}{2}}$ $64^x-1=16^(x/2)$ , find $x$ $x$ ?