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How would I solve: $\log_{16} 32 = x$?

What I know:

  • 16 is the base
  • 32 is the exponent

$$ 32 = 16^x $$

I'm stuck at this point$\ldots$

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  • $\begingroup$ Take logarithms to base 2 of your final equation. $\endgroup$ Commented Aug 12, 2015 at 20:41

4 Answers 4

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$$32=16^x$$

Hint:

$$2^5=2^{4x}$$

Spoiler:

$$4x=5$$ $$\boxed{x=\frac 5 4}$$

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    $\begingroup$ Clear explanation. Thanks! $\endgroup$ Commented Aug 12, 2015 at 20:41
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Hint. Observe that $32 = 16 \cdot 2$, and $16 = 2^4$, so $2 = 16^z$ where $z = $ what?

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  • $\begingroup$ i have to say this is more direct than my answer $\endgroup$
    – pancini
    Commented Aug 12, 2015 at 20:34
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$32 = 16^x$ is the same as $32 = (2^4)^x = 2^{4x}$ and you know that $32 = 2^5$ and so $32 = 2^\color{red}{5} = 2^{\color{red}{4x}}$and so $\color{red}{4x = 5}$ and hence $$x = \frac{5}{4}$$

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First, i want to point out that $x=\log_{16}32$ IS a solution, just not a very useful one since you need to convert to base $e$ or base $10$ to use a calculator usually.

Now that you have $16^x=32$, you can use a log of known base (most commonly, natural log) and plug that in to a calculator to find a value.

$\ln(16^x)=\ln32$

$x\ln16=\ln32$

$x=\frac{\ln32}{\ln16}$

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  • $\begingroup$ It may be better to use $\lg x \equiv \log_2 x$. $\endgroup$
    – Brian Tung
    Commented Aug 12, 2015 at 20:34
  • $\begingroup$ agreed. i was commenting on your answer right when you said that. $\endgroup$
    – pancini
    Commented Aug 12, 2015 at 20:34
  • $\begingroup$ Ok, but my quiz said the answer was $5/4$ $\endgroup$ Commented Aug 12, 2015 at 20:35
  • $\begingroup$ But in fact, your answer is just as direct, provided one uses $\lg x$. $\endgroup$
    – Brian Tung
    Commented Aug 12, 2015 at 20:35
  • $\begingroup$ Can you please show me how to get to 5/4? All of the possible answers were fractions $\endgroup$ Commented Aug 12, 2015 at 20:37

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