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Write 2^5=32 in logarithmic form.
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Can someone please explain, I forget how to do this.
\[x^{a} = b \rightarrow \log_{x} b = a\]
Plus i t is better to use what @iPwnBunnies said.. :)
so \[\log_{2}32=2 \] then?
no
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\[\log_{2}32=5 \]
Hell yeah. B)
Thanks :D
The logarithm operator is used to find exponents. So, you can rewrite stuff with exponents like that.
so you could use log to solve \[\log_{2}x=5 \]
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Yes..
K, got it.
Well, it's used to solve what's on the right of the equal sign, which is supposed to be the exponent. \[2^{x} = 32\] Solve for x. \[\log_{2} 32 = x\]
I see
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