# Logs/Equations

• Oct 6th 2009, 11:33 AM
A Beautiful Mind
Logs/Equations
The problem is: $\displaystyle 4^x-2^{x+1}=3$

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

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

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

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

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

$\displaystyle 2^x-1=0$

$\displaystyle Let 2^x = y.$

$\displaystyle y-1=0$
$\displaystyle y= 1$

$\displaystyle 2^x = 1$

$\displaystyle 2^0=1$

$\displaystyle x = 0$
• Oct 6th 2009, 11:39 AM
e^(i*pi)
Quote:

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

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

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

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

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

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

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

$\displaystyle Let 2^x = y.$

$\displaystyle y-1=0$
$\displaystyle y= 1$

$\displaystyle 2^x = 1$

$\displaystyle 2^0=1$

$\displaystyle x = 0$

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

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

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

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

This is a quadratic in 2^x
• Oct 6th 2009, 11:53 AM
Soroban
Hello, A Beautiful Mind!

Quote:

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

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

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

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

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

• Oct 6th 2009, 12:00 PM
A Beautiful Mind
EDIT: Just saw Soroban's post.

I was trying to use my teacher's method...lol.
• Oct 6th 2009, 12:42 PM
Matt Westwood
Quote:

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

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

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

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

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

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

$\displaystyle 2^x-1=0$

$\displaystyle Let 2^x = y.$

$\displaystyle y-1=0$
$\displaystyle y= 1$

$\displaystyle 2^x = 1$

$\displaystyle 2^0=1$

$\displaystyle x = 0$

Just in case - I don't think it was pointed out that:

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

is actually wrong, should be:

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