The procedure spelled out by wolf_meister will work just fine whether you’re using logs to the base e (natural logs), to the base 10 (common logs), or any other base—as long as you’re consistent!
Either one works. log[sub]7[/sub], log[sub]143[/sub], log[sub]2.6[/sub], log[sub]1/3[/sub] would also work.
You only need to be consistent. If you take, say, log[sub]2.6[/sub], then you have to use 2.6[sup]to the power of something[/sup] at the end, not 10[sup]to the power of something[/sup] or e[sup]to the power of something[/sup], etc.
Let’s spell out the algebraic solution, step by step.
First, we know that (x)[sup][a][sup]b[/sup][/sup] = (x)[sup][ab][/sup]
And so, we want x[sup].05[/sup] wind up being x[sup]1[/sup] which is simply x.
So, converting the decimal to fractions (algebra hates decimals), we have…
x[sup]1/20[/sup]
And to make that x[sup]1[/sup], will need to multiply 1/20 by 20. And to do that, we’ll have to raise both sides of the equation to the 20th power.
.96[sup]20[/sup] = xsup[sup]20[/sup][/sup]
~.442 = x[sup](1/20 * 20) [/sup]
~.442 = x[sup]1[/sup] = x
Hmmmm… upon further reflection… I’ve used up all my math expertise and can’t answer this next question:
Could the answer also be negative .442?
Sure.
Care to explain this? I know if you’ve got something like 4=x^2 then x can be either 2 or -2, but isn’t this analagous to the case of 4=x^1/2? In that case x is going to be 4 and not -4. (-.442)^.05 should be a multiple of i which is not equal to .96.
Right, Snarky_Kong. If x were negative, x[sup]1/20[/sup] would not be a real number.