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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Take logarithms to base 2 of your final equation. – Mark Bennet Aug 12 '15 at 20:41
4 Answers
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Hint. Observe that $32 = 16 \cdot 2$, and $16 = 2^4$, so $2 = 16^z$ where $z = $ what?
Brian Tung
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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 $$x = \frac{5}{4}$$
alkabary
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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.
$\ln(16^x)=\ln32$
$x\ln16=\ln32$
$x=\frac{\ln32}{\ln16}$
pancini
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But in fact, your answer is just as direct, provided one uses $\lg x$. – Brian Tung Aug 12 '15 at 20:35
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Can you please show me how to get to 5/4? All of the possible answers were fractions – Matt Butler Aug 12 '15 at 20:37
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use a calculator. alternatively, solve $2^5=2^{4x}$ by rewriting as $2^5=(2^4)^x$ – pancini Aug 12 '15 at 20:44