1. Logarithms

Hi!

Solve:
1. $2^{x - 1} = 3$

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

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

$x - 1 = \log_2 3$

$x = \log_2 3 + 1$

2. Re: Logarithms

What are you actually asking? If it's correct? Because everything you've done is fine

3. Re: 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

4. Re: 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 " $x= log_2(3)+ 1$" is a perfectly good solution to the equation.

5. Re: Logarithms

Originally Posted by HallsofIvy
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 " $x= log_2(3)+ 1$" is a perfectly good solution to the equation.

But the poster has not solved for x

6. Re: Logarithms

Originally Posted by Unreal
Solve:
1. $2^{x - 1} = 3$
$x = \log_2 3 + 1$
Originally Posted by bjhopper
But the poster has not solved for x
Why in the world would you say that?

What do you mean by a solution?

7. Re: Logarithms

Originally Posted by Plato
Why in the world would you say that?

What do you mean by a solution?

See post 3.I think the OP does not know how to find the numerical value of x

8. Re: Logarithms

Originally Posted by bjhopper
See post 3.I think the OP does not know how to find the numerical value of x
So, is place of "But the poster has not solved for x" you really meant "But the poster has no numerical value for x"?

9. Re: Logarithms

Originally Posted by bjhopper
But the poster has not solved for x
Of course he has. He has x equal to a specific number, $ln_3(2)+ 1$. 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 $\frac{log(2)}{log(3)}+ 1= \frac{0.3010}{0.47712}+ 1= 0.63093+ 1= 1.06093$? Because that is NOT the correct answer- it is only approximately correct.

It is the same as saying that the solutions to $x^2= 2$ are $\sqrt{2}$ and $-\sqrt{2}$, NOT "1.41421" and "-1.41421".

10. Re: Logarithms

Originally Posted by HallsofIvy
Of course he has. He has x equal to a specific number, $ln_3(2)+ 1$. 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 $\frac{log(2)}{log(3)}+ 1= \frac{0.3010}{0.47712}+ 1= 0.63093+ 1= 1.06093$? Because that is NOT the correct answer- it is only approximately correct.

It is the same as saying that the solutions to $x^2= 2$ are $\sqrt{2}$ and $-\sqrt{2}$, NOT "1.41421" and "-1.41421".

You've jumbled a few digits.Iam saying only that the OP should use base 10 0r base e to find the numerical value of x