Expert: Paul Klarreich - 2/7/2009

Question
Hi Paul,

I am having a problem with a logarithm question and was wondering whether you could point me in the correct direction/help me out with it.

I am attempting to solve an equation using quadratics + logarithms.

The question itself is:

solve - 2^(2x+1) - 5(2^x) + 2 = 0

The part I am not quite sure about is how to deal with the power of (2x+1). The following is an example of how I have answered a similar question so that you can follow the steps I am currently taking.

e.g. solve 5^(2x) - 2(5^x) = 3

step 1 - let y = 5^x, becoming
y^2 - 2y = 3

step 2 - convert to quadratic
y^2 - 2y - 3 = 0,
(y - 3)(y + 1) = 0

therefore y = 3 or -1

step 3, sub 5^x back in place of y
5^x = 3 or -1 (as it cannot equal a negative number only accept value 3)

step 4, use logarithms to determine x
5^x = 3
log5^x = log3
x.log5 = log3
x = log3/log5
x = 0.683 (3 d.p)

I hope that gives some kind of outline as to what I am attempting to do, and that you are able to help.

Thanks


Answer
Questioner:   Haydn
Category:  Advanced Math
Private:  No
 
Subject:  Solving logarithms via quadratics
Question:  Hi Paul,

I am having a problem with a logarithm question and was wondering whether you could point me in the correct direction/help me out with it.

I am attempting to solve an equation using quadratics + logarithms.

The question itself is:

solve - 2^(2x+1) - 5(2^x) + 2 = 0

The part I am not quite sure about is how to deal with the power of (2x+1). The following is an example of how I have answered a similar question so that you can follow the steps I am currently taking.

e.g. solve 5^(2x) - 2(5^x) = 3

step 1 - let y = 5^x, becoming
y^2 - 2y = 3

step 2 - convert to quadratic
y^2 - 2y - 3 = 0,
(y - 3)(y + 1) = 0

therefore y = 3 or -1

step 3, sub 5^x back in place of y
5^x = 3 or -1 (as it cannot equal a negative number only accept value 3)

step 4, use logarithms to determine x
5^x = 3
log5^x = log3
x.log5 = log3
x = log3/log5
x = 0.683 (3 d.p)

I hope that gives some kind of outline as to what I am attempting to do, and that you are able to help.

Thanks
........................
Hi, Mozart, Sorry, I mean
Hi, Haydn,

There is nothing wrong with your approach.  In fact, it works quite nicely on your example:

2^(2x+1) - 5(2^x) + 2 = 0   << typo fixed.

Let  y = 2^x, and use exponent laws to simplify:

The terms are:

First:  2^(2x+1) = 2^x 2^x 2 = 2y^2

Second: 5(2^x) = 5y

Now:

2y^2 - 5y + 2 = 0

(2y - 1)(y - 2) = 0

y = 1/2,   y = 2

Now you go back to x, and you don't even need a calculator.

2^x = 1/2,  x = -1

2^x = 2,  x = 1

They even check nicely.

2^(2x+1) - 5(2^x) + 2 = 0
2^(-2+1) - 5(2^-1) + 2 = 0
2^(-1) - 5(2^-1) + 2 = 0
1/2 - 5/2 + 2 = 0, yes.


2^(2x+1) - 5(2^x) + 2 = 0
2^(2+1) - 5(2^1) + 2 = 0
8 - 10 + 2 = 0,  yes.