# Logarithmic Equations

• April 18th 2012, 02:38 PM
Prentz
Logarithmic Equations
Test time again, struggling on a few questions on the practice test and simply looking for some assistance on tackling the equations!
Question 1:
Consider the following logarithmic equations. Solve for x.

A) http://latex.codecogs.com/gif.latex?...2%7D-3-5logx=0

B) http://latex.codecogs.com/gif.latex?...8-log_%7B2%7D6

Question 2:
Consider the following exponential equations. Solve for x.

http://mathhelpforum.com/image/png;b...BJRU5ErkJggg==

Thanks for all your help and support!
• April 18th 2012, 02:48 PM
MrCryptoPrime
Re: Logarithmic Equations
A)

2 log^2(x)-5 log(x)-3 = 0
Substitute u = log(x):
2 u^2-5 u-3 = 0
Solve the quadratic equation by completing the square:

Divide both sides by 2:
u^2-(5 u)/2-3/2 = 0
u^2-(5 u)/2 = 3/2
u^2-(5 u)/2+25/16 = 49/16
Factor the left hand side:
(u-5/4)^2 = 49/16
Take the square root of both sides:
abs(u-5/4) = 7/4
Eliminate the absolute value:
u-5/4 = -7/4 or u-5/4 = 7/4
u = -1/2 or u-5/4 = 7/4
Substitute back for u = log(x):
log(x) = -1/2 or u-5/4 = 7/4
Cancel logarithms by taking exp of both sides:
x = 1/sqrt(e) or u-5/4 = 7/4
x = 1/sqrt(e) or u = 3
Substitute back for u = log(x):
x = 1/sqrt(e) or log(x) = 3
Cancel logarithms by taking exp of both sides:
x = 1/sqrt(e) or x = e^3
Now test that these solutions are appropriate by substitution into the original equation:
Check the solution x = 1/sqrt(e):
2 log^2(x)-5 log(x)-3 => -3-5 log(1/sqrt(e))+2 log^2(1/sqrt(e)) = 0
So the solution is correct.
Check the solution x = e^3:
2 log^2(x)-5 log(x)-3 => -3-5 log(e^3)+2 log^2(e^3) = 0
So the solution is correct.
Thus, the solutions are:
x = 1/sqrt(e) or x = e^3

B)

(log(x))/(log(2)) = (log(18))/(log(2))-(log(6))/(log(2))
Divide both sides by 1/(log(2)):
log(x) = log(2) ((log(18))/(log(2))-(log(6))/(log(2)))
Cancel logarithms by taking exp of both sides:
x = 3
Now test that this solution is appropriate by substitution into the original equation:
Check the solution x = 3:
(log(x))/(log(2)) => (log(3))/(log(2)) = (log(3))/(log(2)) ~~ 1.58496
(log(18))/(log(2))-(log(6))/(log(2)) => -(log(6))/(log(2))+(log(18))/(log(2)) = (log(3))/(log(2)) ~~ 1.58496
So the solution is correct.
Thus, the solution is:
x = 3

Basically, if you see a quadratic pattern like with A), e.g. in the form ax^2+x+c, where c=some constant, then solve using quadratic equation or completing square. Best to make sure you have your log rules down too:
Mathwords: Logarithm Rules

Hope that helps? Best of luck! ("problem 2's" picture appears to not be working...not displaying correctly?)
• April 18th 2012, 05:12 PM
skeeter
Re: Logarithmic Equations
Quote:

Originally Posted by Prentz

$\log_2{x} = \log_2{18} - \log_2{6}$

$\log_2{x} = \log_2\left(\frac{18}{6}\right)$

$\log_2{x} = \log_2\left{3}$

$x = 3$

... if you meant to post more after this equation (question 2 ?), it did not show.
• April 19th 2012, 04:49 PM
Prentz
Re: Logarithmic Equations
Thanks for the help guys, is there an easier way to express question 1? It confused me slightly.
And as for question 2:
Consider the following exponential equations. Solve for x.
Attachment 23661
• April 19th 2012, 04:50 PM
Prentz
Re: Logarithmic Equations
Any help you can give on that question 2 and clearing up question 1 would be great!
• April 19th 2012, 05:26 PM
HallsofIvy
Re: Logarithmic Equations
Do you know anything about the "laws of exponents"?

$a^xa^y= a^{x+y}$
$(a^x)^y= a^{xy}$
• April 19th 2012, 06:13 PM
highvoltage
Re: Logarithmic Equations
Laws of exponents.
• April 19th 2012, 07:22 PM
highvoltage
Re: Logarithmic Equations
• April 19th 2012, 10:53 PM
a tutor
Re: Logarithmic Equations
Quote:

Originally Posted by Prentz

Yet another solution...why?

$2(\log x)^2-5\log x -3=0$

$(2\log x +1)(\log x-3)=0$

$\log x = -\frac{1}{2}$ or $\log x =3$

$x=e^{-1/2}$ or $x=e^3$