Math Help - Logarithm Equation

1. Logarithm Equation

Solve $\lg(4^p - 4) - p\lg2 = \lg3$
I am supposed to do corrections for this question but I do not know how.
The correct answer is $p = 2$
Here are the steps I have done:
$\lg(4^p - 4) - p\lg2 = \lg3$
$\lg(4^p - 4) - \lg2^p = \lg3
$

$\lg\frac{4^p - 4}{2^p} = \lg3$
$\frac{4^p - 4}{2^p} = 3$
$4^p - 4 = 2^p \times 3$
$\frac{2^{2p} - 4}{2^p} = 3$
$\frac{2^{2p}}{2^p} - \frac{4}{2^p} = 3$
$2^p - 2^{2 - p} = 3$
$2^p(1 - 2^{2 - 1}) = 3$
$2^p = -3$

Any help would really be appreciated, thank you very much in advance!

2. $2^p - 2^{2 - p} = 3$

This line cannot become:

$2^p(1 - 2^{2 - 1}) = 3$

it becomes:

$2^p(1 - 2^{2 - 2p}) = 3$

$2^{p} - \dfrac{4}{2^p} = 3$

From here, you can use a substitution of the form y = 2^p

$y - \dfrac{4}{y} = 3$

Now, can you continue?

EDIT: Oops, sorry mr fantastic

3. Originally Posted by caramelcake
I am supposed to do corrections for this question but I do not know how.
The correct answer is $p = 2$
Here are the steps I have done:
$\lg(4^p - 4) - p\lg2 = \lg3$
$\lg(4^p - 4) - \lg2^p = \lg3
$

$\lg\frac{4^p - 4}{2^p} = \lg3$
$\frac{4^p - 4}{2^p} = 3$
$4^p - 4 = 2^p \times 3$
$\frac{2^{2p} - 4}{2^p} = 3$
$\frac{2^{2p}}{2^p} - \frac{4}{2^p} = 3$
$2^p - 2^{2 - p} = 3$
$2^p(1 - 2^{2 - 1}) = 3$ Mr F says: This line is wrong. Expand it out to see why. $2^p 2^{1} \neq 2^{2-p}$.
$2^p = -3$

Any help would really be appreciated, thank you very much in advance!
$4^p - 4 = 2^p \times 3$

$\Rightarrow (2^{p})^2 - 3 \times 2^p - 4 = 0$

$\Rightarrow (2^p - 4)(2^p + 1) = 0$

etc.

4. Originally Posted by caramelcake
I am supposed to do corrections for this question but I do not know how.
The correct answer is $p = 2$
Here are the steps I have done:
$\lg(4^p - 4) - p\lg2 = \lg3$
$\lg(4^p - 4) - \lg2^p = \lg3
$

$\lg\frac{4^p - 4}{2^p} = \lg3$
$\frac{4^p - 4}{2^p} = 3$
$4^p - 4 = 2^p \times 3$
$4^p= (2^2)^p= (2^p)^2$ so you can think of $4^p- 3(2^p)- 4= (2^p)^2- 3(2^p)- 4= 0$ as a quadratic equation for $2^p$. As Unknown008 (who is that masked man?) suggested, let $y= 2^p$ and you get $y^2- 3y- 4= 0$. Solve that quadratic equation for y and then solve $2^p= y$ for p.

$\frac{2^{2p} - 4}{2^p} = 3$
$\frac{2^{2p}}{2^p} - \frac{4}{2^p} = 3$
$2^p - 2^{2 - p} = 3$
$2^p(1 - 2^{2 - 1}) = 3$
$2^p = -3$

Any help would really be appreciated, thank you very much in advance!

5. I'm not masked, I'm Jerry

6. Thank you everybody! I am very touched by the number of responses to my question. Sorry for not replying earlier, I am not really feeling well today.

7. It's okay, we're glad to help

8. Originally Posted by Unknown008
$2^p - 2^{2 - p} = 3$
at this point either p or 2-p have to be 0, since the difference of 2 powers of 2 give an odd number only if the power of one is 0. also p cannot be zero since then 2-p=2 which gives -3.