Logarithm question

**yorkey** · May 7th 2011, 02:11 AM

If the sum is 2^x = 8, how do I work out x?

When I use my calculator, I press 'log', then 2 and then in brackets (8), so it looks like this "log2(8)". Why do I get the answer of 2.4082399.....

**earboth** · May 7th 2011, 02:19 AM

Originally Posted by yorkey

If the sum is 2^x = 8, how do I work out x?

When I use my calculator, I press 'log', then 2 and then in brackets (8), so it looks like this "log2(8)". Why do I get the answer of 2.4082399.....

There are a lot of different ways to solve this equation. I'm going to show you 2 of them:

1. Change 8 into a power of 2:

$2^x = 8~\implies~2^x=2^3$

Two powers of the same base are equal if the exponents are equal too.

2. Use logarithms (but correct) and the base-change-formula:

$2^x = 8~\implies~x=\log_2(8) ~\implies~x=\dfrac{\log(8)}{\log(2)} = \dfrac{\ln(8)}{\ln(2)}$

**pickslides** · May 7th 2011, 02:21 AM

Using logs you can say

$\displaystyle 2^x = 8$

$\displaystyle \log_22^x = \log_28$

$\displaystyle x = \log_22^3$

$\displaystyle x = 3\log_22$

$\displaystyle x = 3\times 1$

$\displaystyle x = 3$

Seems like the long way home though, do you agree?

**yorkey** · May 7th 2011, 02:26 AM

Easy peazy - now that I know what to do, and my questions won't get more complicated than this. Thank yuo!

**Auri** · May 7th 2011, 08:36 AM

By pressing log2(8) you're not saying base 2, index 8. Calculators by default use the 10 base. You'll have to use the base change formula, which is log of the index (by default base 10)/ log of the base (by default base 10).

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