# Math Help - Logarithmic Equations

1. ## 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) $2(logx)^{2}-3-5logx=0$

B) $log_{2}x=log_{2}18-log_{2}6$

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

Thanks for all your help and support!

2. ## 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
Add 3/2 to both sides:
u^2-(5 u)/2 = 3/2
Add 25/16 to both sides:
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
Add 5/4 to both sides:
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
Add 5/4 to both sides:
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?)

3. ## Re: Logarithmic Equations

Originally Posted by Prentz
B) $log_{2}x=log_{2}18-log_{2}6$
$\log_2{x} = \log_2{18} - \log_2{6}$

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

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

$x = 3$

... if you meant to post more after this equation (question 2 ?), it did not show.

4. ## 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.

5. ## Re: Logarithmic Equations

Any help you can give on that question 2 and clearing up question 1 would be great!

6. ## Re: Logarithmic Equations

Do you know anything about the "laws of exponents"?

$a^xa^y= a^{x+y}$
$(a^x)^y= a^{xy}$

7. ## Re: Logarithmic Equations

Laws of exponents.

9. ## Re: Logarithmic Equations

Originally Posted by Prentz

A) $2(logx)^{2}-3-5logx=0$
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 $x=e^3$