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Discover $x$ if $2^{16^x} = 16^{2^10}$.

If $x = 0$ we accept $2=sixteen$.

$eight$ times or $(ii^3)$ difference and if $x = 1$ we take $65536=256$.

$256$ or $2^eight$ divergence. I tin't see anything useful in figuring this out.

I have a feeling $x$ must be a fraction. I know simply know well-nigh logarithm. I oasis't learned annihilation most the number $due east$ or $ln$ yet.

  • algebra-precalculus

Blue's user avatar

Blue

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asked Nov thirteen, 2020 at nineteen:11

Edmund's user avatar

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half-dozen

  • $\begingroup$ take log base two, y'all'll go $2^{4x}$ = $ii^{x+2}$ $\endgroup$

    Nov thirteen, 2020 at 19:xvi

  • $\begingroup$ @Aditya_math so yous would desire to take $\log_2$ a second time $\endgroup$

    November 13, 2020 at 19:19

  • $\begingroup$ @Henry yeah, or simply we could just compare powers which is technically the same thing! $\endgroup$

    Nov 13, 2020 at 19:xx

  • $\begingroup$ Lets grouping the $x$'due south on one side and the numbers on the other side, it is clear that $ten+ten=\frac{216}{162}\implies x=\frac 23$. Ok, I'll just be leaving now... $\endgroup$

    Nov thirteen, 2020 at twenty:44

  • $\begingroup$ @zwim how did yous derive $\frac{216}{162}$? $\endgroup$

    Nov xiii, 2020 at 23:49

three Answers 3

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$xvi^{2^x}=(ii^four)^{2^ten}=2^{4\cdot ii^x}=two^{2^2\cdot two^x}=two^{2^{x+2}}$. And so $$ii^{16^x}=16^{2^x}\iff ii^{16^10}=2^{2^{x+ii}}\iff$$ $$\iff 16^x=2^{x+2}\iff 2^{4x}=ii^{x+2} \iff 4x=x+2 \iff x=\frac{2}{iii}. $$

answered November 13, 2020 at nineteen:16

Tito Eliatron's user avatar

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$$ii^{xvi^x} = 2^{(ii^iv)^x} = two^{2^{4x}}$$

$$16^{ii^x} = (2^4)^{two^x} = two^{4\times 2^x} = 2^{2^{x+2}}$$

Now equate the two expressions.

answered November 13, 2020 at 19:xv

player3236's user avatar

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If you have the $\log_2$ of both sides you stop up with \brainstorm{align*} \log_2(two^{16^10}) &= 16^x \log_2(two) = 16^x = 2^x\cdot8^x\\ \log_2(xvi^{2^10}) &= two^10\log_2{sixteen} = 2^ten\cdot4 \end{align*} From this you go that $two^x\cdot four = ii^x \cdot eight^x \implies eight^ten = 4 \implies x =\log_{8}(4) = ii/3.$

($\log_8(iv) = \log_8(ii^two) = 2\log_8(2) = 2\cdot1/3$)

answered Nov 13, 2020 at 19:21

Matt Werenski's user avatar

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