# Math Help - Logs/Equations

1. ## Logs/Equations

The problem is: $4^x-2^{x+1}=3$

Here's the work I've tried doing on it:

$4^x-2^{x+1}=3$

I split the $2^{x+1}$ up into $-2^x+2^1$.
Then I...

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

$4^x-2^x -1= 0$

$2^x-1=0$

$Let 2^x = y.$

$y-1=0$
$y= 1$

$2^x = 1$

$2^0=1$

$x = 0$

2. Originally Posted by A Beautiful Mind
The problem is: $4^x-2^{x+1}=3$

Here's the work I've tried doing on it:

$4^x-2^{x+1}=3$

I split the $2^{x+1}$ up into $-2^x+2^1$.
Then I...

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

$4^x-2^x -1= 0$

$2^x-1=0$ me: (Different bases mean you can't take it away like that )

$Let 2^x = y.$

$y-1=0$
$y= 1$

$2^x = 1$

$2^0=1$

$x = 0$
$4^x-2^{x+1}=3$

$-2^{x+1} = -(2^{x+1}) = -2^x-2$

The key to this question is recognising that 4 can be written as 2^2

$2^{2x}-2^x-5=0$

This is a quadratic in 2^x

3. Hello, A Beautiful Mind!

Solve for $x\!:\;\;4^x-2^{x+1}\:=\:3$

We have: . $(2^2)^x - 2\!\cdot\!2^x - 3 \:=\:0 \quad\Rightarrow\quad 2^{2x} - 2\!\cdot\!2^x - 3 \:=\:0$

Factor: . $\left(2^x + 1\right)\left(2^x - 3\right) \:=\:0$

Then: . $\begin{array}{ccccccc}2^x+1 \:=\:0 & \Rightarrow & 2^x \:=\:1 & &\text{no real roots} \\
2^x - 3 \:=\:0 & \Rightarrow & 2^x \:=\:3 & \Rightarrow & x \:=\:\log_2(3)
\end{array}$

Therefore: . $x \:=\:\log_2(3)\:\text{ or }\:\frac{\ln(3)}{\ln(2)}$

4. EDIT: Just saw Soroban's post.

I was trying to use my teacher's method...lol.

5. Originally Posted by A Beautiful Mind
The problem is: $4^x-2^{x+1}=3$

Here's the work I've tried doing on it:

$4^x-2^{x+1}=3$

I split the $2^{x+1}$ up into $-2^x+2^1$.
Then I...

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

$4^x-2^x -1= 0$

$2^x-1=0$

$Let 2^x = y.$

$y-1=0$
$y= 1$

$2^x = 1$

$2^0=1$

$x = 0$
Just in case - I don't think it was pointed out that:

$- 2^{x+1}= -2^x+2^1$

is actually wrong, should be:

$- 2^{x+1} = - (2^x)(2^1)$