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I have been trying to rearrange an equation involving a log function, as shown below: $$ log(a(b-1)N/s^{-b} + 1) = y. $$ I am trying to make the N term the subject of this equation, but am having some trouble in doing this. From using a Symbolab equation rearranger, I got this expression: $$ N = 10^y s^{-b}-s^{-b}/a(-1+b) $$ , but I'm not too sure if this is correct.

Can someone please provide some guidance on this matter? Any form of help would be appreciated.

Cheers, Al

Matti P.
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jon_doe
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    If you are not sure then plug it in and see if it checks out. – Michal Adamaszek Sep 10 '20 at 12:05
  • The expressions are a bit vague (even though I tried to make them better) - can you check them and see which terms are supposed to be divided by which terms? Then I could check if it works. Thanks. – Matti P. Sep 10 '20 at 12:15

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Since N is "inside" a logarithm, you need to reverse that by taking the inverse function, the exponential function, to both sides. You don't say what the base of the logarithm is, whether it is the "common logarithm" (base 10) or "natural logarithm" (base e) so I will just call it "a".

$\log\left(\frac{a(b- 1)N}{s^{-b}+ 1}\right)= y$

Taking the exponential of both sides
$\frac{a(b- 1)N}{s^{-b}+ 1}= a^y$ (Your "symbolab equation rearranger" is assuming "common logarithm", base 10. Is that correct?) Now, divide both sides by a(b- 1) and multiply both sides by $s^{-b}+ 1$ to get $S= \frac{(s^{-b}+ 1)a^y}{a(b-1)}$.

I am assuming that your "$a(b−1)N/s^−b+1$" was actually "$\frac{a(b-1)N}{s^{-b}+ 1}$" rather than $\frac{a(b- 1)N}{s^{-b}}+ 1$ which is what you actually wrote. Is that correct?

user247327
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  • Well @user247327 my expression is actually the one with the plus one at the end (the last equation in the second last sentence in your reply. The logarithm used is a base 10. Sorry for the confusion. – jon_doe Sep 10 '20 at 23:39