Hey, it has been a while since I have used logs, so I am having trouble solving this math problem: 20 = 3log(x+1) Thanks!
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Originally Posted by christiangirl Hey, it has been a while since I have used logs, so I am having trouble solving this math problem: 20 = 3log(x+1) Thanks! assuming log is base 10 ... $\displaystyle 20 = 3\log(x+1)$ $\displaystyle 20 = \log(x+1)^3$ $\displaystyle (x+1)^3 = 10^{20}$ $\displaystyle x+1 = 10^{\frac{20}{3}}$ $\displaystyle x = 10^{\frac{20}{3}} - 1 $
oh sorry it is base 2...
no matter, the procedure is given above, apply it anyway.
is it any different if it is base 2?
oh I see I use 2 instead of 10. Thanks!
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