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?)

Re: Logarithmic Equations

Quote:

Originally Posted by

**Prentz**

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

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

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

$\displaystyle x = 3$

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

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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

Re: Logarithmic Equations

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

Re: Logarithmic Equations

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

$\displaystyle a^xa^y= a^{x+y}$

$\displaystyle (a^x)^y= a^{xy}$

Re: Logarithmic Equations

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Re: Logarithmic Equations

Re: Logarithmic Equations

Quote:

Originally Posted by

**Prentz**

Yet another solution...why?

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

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

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

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