# math question, solving for x when i know the exponent and the answer

I have the following

.96=x^.05

what’s the manipulation to solve for x?

IOW, .96^(?)=x

logs

Or x = .96[sup]1/.05[/sup]

That’s taking the .05th root of each side.

JoeyP didn’t show the entire procedure:

If .96=x^.05 then by taking logs of both sides we have:

log(.96) = .05 * log(x)

log(.96)
------------ = log(x)
.05

-.01773
------------ = log(x)
.05

-0.35458 = log(x)

x = .70147

Hmm seems that is a different answer from what JoeyP has.

Yeah, you used log base 10 in taking the log of .96 and the natural log of x in your last step.

Damn I was using Excel to find those logs. I always thought Excel defaulted to natural logs. Seems it defaults to base 10 logs. :smack:

SO… reworking the last step

-0.35458 = log(x)

x = 10^-0.35458

x = .442002

and JoeyP’s answer is

.96^20 which equals

.442002 which agrees with the “revised” answer.

Both your methods are correct, however wolf_meister you must have made some calculation error in the final step since the correct answer should be 0.442002 which is

a) 0.96^1/0.05
or
b) 10^log(x)

Your last step is wrong Wolf.
if -0.35458 = log(x)
then x = 10^(-.35458) = .442

I can see what you did though - you used base 10 logs to start, then you switched to natural logs at the end.

Wow. My record for simulposts.

Good work Snarky.

As I was typing my correction, you already posted yours.

I was using the log function in Excel which yielded natural logs* in earlier versions*. Now it seems that “log” yields base 10 logs.

Ironically, I should have used my own logarithm calculator at
http://www.1728.com/logrithm.htm

:smack:

The solution of x^a = b does not require logs. The easy answer is x = b^(1/a). Now if the unknown x is part of an exponent, then you’ll have to resort to logs. That’s how you can bring x “down” a level to solve.

Nightime
and it seems I set a personal record for 2 “smack-the-head” postings in the same thread. Heck I was amazed that I got it in there before everyone else.

Log, shmog. Just raise both sides of the equation to the 20th power, which is equivalent to Joey P’s second post, only more simply explained.

wolf_meister: Excel is designed to be used in offices. It generally makes scientific and technical calculations as difficult as possible – so why would you think it would default to natural logs? =) (Not that I can imagine many business applications for logarithms anyway, but still.)

I like how you can’t use superscript or subscript in the graphing tool (or in certain places in Powerpoint, IIRC). Try making a graph called ‘Dissociation Constants of Halogen-Substituted Derivatives of Butanoic Acid (CH[sub]3[/sub]CH[sub]2[/sub]CH[sub]2[/sub]COOH)’. Polyatomic ions are also impossible (something like ClO[sub]4[/sub][sup]-[/sup], only the superscript and subscript are aligned vertically. If you want that to look right, you have to use Equation Editor.

I don’t know off hand whether it does or not, but if it does, I’d guess it’s because that’s the easiest kind for the computer to calculate. As far as I know, the computer’s algorithm for finding logs to other bases (like common logs) would involve finding natural logs and converting.

The OP’s equation could also be written as .96 = x[sup]1/20[/sup], which makes it more obvious why Bryan Ekers’s simple solution is correct.

It’s no harder to find one than the other, and even if it were, the conversion process consists simply of multiplying by a constant. The constant is log[sub]10[/sub] e = 0.434294. To convert the other way, you multiply by log[sub]e[/sub] 10 = 2.30259 = 1 / 0.434294.

I take that back. It can be slightly easier to calculate log[sub]e[/sub] than log[sub]10[/sub], but only because you can omit a single multiplication if you’re using a power series.

As I said, the Excel default for logs was natural logs but I suppose they changed it in time for me to make my miscalculation. Instead of using their anti-log function (which would be exp ???) I figured I’d take the easy way (and incorrect way) and raise ‘e’ (I typed out 2.718281828) and to the -.35458th power. It would have been easier (and correct) to have used 10.

In case you are wondering, my log calculator (and the others) are correct. The one thing I like about programming is that you have to get your calculations correct just once. Then you don’t have to worry about which logs to use, etc.

I also think Bryan Ekers answer / explanation is the best. Of course JoeyP was first.

So, in general, when solving problems such as the OP, do you use ln? or log? And, if it’s ln (natural log) how do we know this? If ln, then how did “e” enter into the problem…and, why doesn’t base 10 work? It seems counterintuitive that base 10 wouldn’t work…when our whole number system is based on powers of ten. - Jinx