Solving for $x$ using exponential log laws

Question

For

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

I moved all the $x$ to left side, used got rid of log and got

$x-(x-1)^2 - (x+1) = 4$

Simplyifing I get

$x^2-2x=4$

The answer should be $x = 4$ (I checked on wolfram alpha) Help please?

Answer 1

Here are the steps I would use:

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

$\log_2(x) + \log_2\left((x-1)^2\right) = \log_2\left(2^2\right) + \log_2(2x+1)$

That step turns each term into a base $2$ logarithm. Now they can be combined using the product rule for logs:

$\log_2\left(x(x-1)^2\right) = \log_2\left(4(2x+1)\right)$

Now you can get rid of the logarithms, and you get:

$x(x-1)^2 = 4(2x+1)$,

or

$x^3 - 2x^2 - 7x - 4 = 0$

This can be factored as $(x-4)(x + 1)^2=0$, which has solutions $x=4$ and $x=-1$. However, the second of these solutions is extraneous, which you can see by trying to plug it into the original equation.

Does that all makes sense?

Answer 2

Using $\log a+\log b=\log(ab), m\log c=\log(c^m)$

We have $\log_2x(x-1)^2=\log_2(2^2)+\log_2(2x+1)=\log_24(2x+1)$

$\implies x(x-1)^2=4(2x+1)$

Answer 3

So many mistakes here. Your original intent seemed to be a good one. That should give

$$\log_2x+\log_2(x-1)-\log_2(2x+1)=2$$

What it looks like you did from here was to raise $2$ to the power of each term to get

$$x+(x-1)^2-(x+1)=4$$

It looks like you dropped the $2$ in the last log term and miscopied the problem here with a minus where you had a plus. In addition to that, the method is incorrect. You must use the properties $\log a+\log b=\log ab$ or $n^{a+b}=n^an^b$. It's probably best to use properties of logarithms to combine everything on the left into a single logarithm. This results in

$$\log_2\dfrac{x(x-1)^2}{2x+1}=2$$ $$\dfrac{x(x-1)^2}{2x+1}=4$$ $$x^3-2x^2+x=8x+4$$ $$x^3-2x^2-7x-4=0$$ $$(x-4)(x+1)^2=0$$

Since the log of a negative number is undefined, the only solution that works is $x=4$.