Find x in this equation. I know what x is, but I don't know the steps in getting it!!

auonline · #1 $Old$ 08-22-2008, 07:03 PM

Hi all
We have our maths exam on tuesday and I am doing revision and there is a question on this sheet we have that I can't figure out the process of answering it for.

I have to find x, and I figured it out by guess + checking on the calculator, but how do I do it algebraically?
$4^x+3*4^x=1$

What is the steps needed to find the answer?

Any help is appreciated
Lachlan

EDIT:
Nevermind I got it

ended up being something simple which I overlooked...
I forgot that $4^x+3*4^x=4*4^x$ so I understood it and divided both sides by 4, and then went from there

Chop Suey · #2 $Old$ 08-22-2008, 07:19 PM

To solve this, you must use logarithms. Remember some logarithm properties

$\log_a{(x \cdot y)} = \log_a{x} + \log_a{y}$

$\log_a{\left(\frac{x}{y}\right)} = \log_a{x} - \log_a{y}$

$\log_a{(x)^n} = n\log_a{x}$

$\log_a{a} = 1$

$\frac{1}{a} = a^{-1}$

====================

Back to your question.

Given: $4^x + 3 * 4^x = 1$

Pull 4^x common factor: $4^x(1+3) = 1$

Divide by 4 on both sides: $4^x = \frac{1}{4} = 4^{-1}$

Take the log of base 4 on both sides: $\log_4{4^x} = \log_4{4^{-1}}$

Carry down x and -1: $x = -1$

Remember that $\log_4{4} = 1$