# Solving Logarithmic Equations Help. Any Aid Appreciated

• September 20th 2010, 01:48 AM
BinaryBoy
Solving Logarithmic Equations Help. Any Aid Appreciated
Hi everyone,

This forum provides such excellent help that I am back. :D

This time I am attempting to solve basic logarithmic equations. As I was working my way through the questions, a couple of them stumped me.

Here they are:

Solve for x:

1) 5 logx (625) = 10

2) -log3x -1 (1/32) = 5

So, x, and 3x -1 are the respective bases of the logarithm in the above examples. In the first question, I started by dividing both sides by "5" to get logx 625 = 2. And then I simplified to get 4 logx 5 = 2. I divided both sides by 4 again and achieved: logx 5 = 2/4 = 1/2

But from there I am not sure how to go. And I was on an utterly wrong tangent for question 2.

Any help at all will be much appreciated! :D

Thanks,

Nathaniel
• September 20th 2010, 01:52 AM
Prove It
$5\log_x{(625)} = 10$

$\log_x{(625)} = 2$

$\log_x{(25^2)} = 2$

$2\log_x{(25)} = 2$

$\log_x{(25)} = 1$

$x^{\log_x{(25)}} = x^1$

$25 = x$.
• September 20th 2010, 01:55 AM
Prove It
$-\log_{3x-1}\left(\frac{1}{32}\right) = 5$

$\log_{3x - 1}\left[\left(\frac{1}{32}\right)^{-1}\right] = 5$

$\log_{3x - 1}(32) = 5$

$\log_{3x - 1}(2^5) = 5$

$5\log_{3x - 1}(2) = 5$

$\log_{3x - 1}(2) = 1$

$(3x - 1)^{\log_{3x - 1}(2)} = (3x - 1)^1$

$2 = 3x - 1$

$3 = 3x$

$x = 1$.
• September 20th 2010, 02:39 AM
BinaryBoy
Thank you so much, Prove It! :D

Also, two more things: with the first equation, when you reach logx 25 = 1, why is it that no further simplification takes place upon the 25 numeral? I know x = 25 is the correct answer, but I am just curious as to why no further simplification takes place.

In the second equation, how come you don't divide 5 by -1 to yield -5?

Cheers,

Nathaniel
• September 20th 2010, 09:33 AM
Soroban
Hello, BinaryBoy!

If you use the identity: . $\log_b(N) \:=\:x \quad\Longleftrightarrow\quad b^x \:=\:N$

. . you can avoid a lot of manipulation.

Quote:

$1)\;5\log_x(625) \:=\: 10$

We have: . $\log_x(625) \:=\:2$

Then: . $x^2 \:=\:625 \quad\Rightarrow\quad x \:=\:\pm25$

Since the base of a log must be positive: . $\boxed{x \,=\,25}$

Quote:

$2)\; -\log_{3x -1}\!\left(\dfrac{1}{32}\right) \:=\: 5$

We have: . $\log_{3x-1}\!\left(\dfrac{1}{32}\right) \:=\:-5$

Then: . $(3x-1)^{-5} \:=\:\dfrac{1}{32} \;=\;\dfrac{1}{2^5}$

So we have: . $(3x-1)^{-5} \;=\;2^{-5}$

Therefore: . $3x-1 \:=\:2 \quad\Rightarrow\quad 3x \:=\:3 \quad\Rightarrow\quad \boxed{x \:=\:1}$

• September 20th 2010, 02:53 PM
BinaryBoy
Ah. Thanks a heap, Soroban. :)
• September 21st 2010, 02:31 AM
HallsofIvy
Quote:

Originally Posted by BinaryBoy
Thank you so much, Prove It! :D

Also, two more things: with the first equation, when you reach logx 25 = 1, why is it that no further simplification takes place upon the 25 numeral? I know x = 25 is the correct answer, but I am just curious as to why no further simplification takes place.

He could have said $log_x(25)= log_x(5^2)= 2log_x(5)= 1$ so that $log_x(5)= \frac{1}{2}$ but does that really simplify anything?

$log_x(25)= 1$ leads to 25= x^1 and $log_x(5)= \frac{1}{2}$ leads to 5= x^{1/2} which then gives x= 25.

The crucial point is: $log_a(b)= c$ if and only if $a^c= b$.

Quote:

In the second equation, how come you don't divide 5 by -1 to yield -5?
If you divide both sides of $-log_{3x-1}(\frac{1}{32})= 5$ by -5, you get $log_{3x-1}(\frac{1}{32})= -5$ which (again changing $log_a(b)= c$ to $a^c= b$ becomes $(3x-1)^{-5}= \frac{1}{32}$ and probably the best thing to do would be to recognize that $(3x-1)^{-5}= \frac{1}{(3x-1)^5}= \frac{1}{32}$ so that $(3x-1)^5= 32= 2^5$.

Instead, seeing that there was already a fraction inside the logarithm, Prove It chose to use the law of logarithms, taking the -1 inside the logarithm to "cancel" that fraction: $-log_{3x-1}\left(\frac{1}{32}\right)= log_{3x-1}\left(\left(\frac{1}{32}\right)^{-1}\right)= log_{3x-1}(32)$

Once again, the crucial point for all these problems is that $log_a(b)= c$ is exactly the same as $a^c= b$.
Quote:

Cheers,

Nathaniel