Hi!

Solve:

1. $\displaystyle 2^{x - 1} = 3$

$\displaystyle \log_2 2^{x - 1} = \log_2 3$

$\displaystyle (x - 1)\log_2 2 = \log_2 3$

$\displaystyle x - 1 = \log_2 3$

$\displaystyle x = \log_2 3 + 1$

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- Apr 29th 2013, 12:26 AMUnrealLogarithms
Hi!

Solve:

1. $\displaystyle 2^{x - 1} = 3$

$\displaystyle \log_2 2^{x - 1} = \log_2 3$

$\displaystyle (x - 1)\log_2 2 = \log_2 3$

$\displaystyle x - 1 = \log_2 3$

$\displaystyle x = \log_2 3 + 1$ - Apr 29th 2013, 02:37 AMProve ItRe: Logarithms
What are you actually asking? If it's correct? Because everything you've done is fine :)

- Apr 29th 2013, 06:33 AMbjhopperRe: Logarithms
Hi unreal,

Why use base 2 and not base 10 or base e.Othewise you will have to change base to solve for x - Apr 29th 2013, 06:48 AMHallsofIvyRe: Logarithms
I disagree with that. In order to get a specific

*numerical*value for x, yes, you would have to change to base 10 or base e.

But "$\displaystyle x= log_2(3)+ 1$" is a perfectly good**solution**to the equation. - Apr 29th 2013, 09:09 AMbjhopperRe: Logarithms
- Apr 29th 2013, 09:15 AMPlatoRe: Logarithms
- Apr 29th 2013, 10:48 AMbjhopperRe: Logarithms
- Apr 29th 2013, 11:00 AMPlatoRe: Logarithms
- Apr 29th 2013, 11:44 AMHallsofIvyRe: Logarithms
Of course he has. He has x equal to a specific number, $\displaystyle ln_3(2)+ 1$. Perhaps you don't like the way the number is written, but that has nothing to do with "solving". Are you saying that he should solve for $\displaystyle \frac{log(2)}{log(3)}+ 1= \frac{0.3010}{0.47712}+ 1= 0.63093+ 1= 1.06093$? Because that is NOT the correct answer- it is only approximately correct.

It is the same as saying that the solutions to $\displaystyle x^2= 2$ are $\displaystyle \sqrt{2}$ and $\displaystyle -\sqrt{2}$, NOT "1.41421" and "-1.41421". - Apr 29th 2013, 01:19 PMbjhopperRe: Logarithms