1. ## Solving Logarithmic Equation

Hi,

Just wondering is someone could have a look at this equation and check my answer for me?

Find the real solutions of log3(10x^2 - x - 2) = 2 + 2log3(x)

My answers came out to x = 2, -1 and I have said that x=2 is the only real solution.

In terms of my logic for why -2 is the only real solution, I came to this because I know we can't have the log of a negative number as a real solution, so 2log3(x) would mean x must be >0. Can anyone help me get a better understanding of how you should first decide what the real solutions must be and then, how to set this out on paper showing my reasoning?

Thanks

3. ## Re: Solving Logarithmic Equation

Originally Posted by jpritch422
do you mean log[3(10x^2 - x - 2)] or (10x^2 - x - 2)(log 3) ?
I think the OP means $\log_3(10x^2-x-2) = 2+2\log_3(x)$

Originally Posted by andy000
Hi,

Just wondering is someone could have a look at this equation and check my answer for me?

Find the real solutions of log3(10x^2 - x - 2) = 2 + 2log3(x)

My answers came out to x = 2, -1 and I have said that x=2 is the only real solution.

In terms of my logic for why -2 is the only real solution, I came to this because I know we can't have the log of a negative number as a real solution, so 2log3(x) would mean x must be >0. Can anyone help me get a better understanding of how you should first decide what the real solutions must be and then, how to set this out on paper showing my reasoning?

Thanks
You are correct that 2 is the only real solution. To describe it on paper, I would write out the steps you used to find the two solutions. Then, I would say that since $\log_3(x)$ is not defined over the reals for negative values of $x$, $-1$ cannot be a solution.

4. ## Re: Solving Logarithmic Equation

Understood. Edited my post.