# Some Logarithm Questions

• December 19th 2006, 06:22 PM
SportfreundeKeaneKent
Some Logarithm Questions
Here are some log questions that I haven't been able to solve. The problem that I'm having is that the base is different in some of them which is throwing me off.
By the way, the first question has a little typo. The right side of the equation should say 2log9(x-3) instead of 2log5(x-3)

http://img133.imageshack.us/img133/2...0132021kf7.png

If you can't see that picture then here's a link:
http://img133.imageshack.us/img133/2...0132021kf7.png
• December 19th 2006, 07:18 PM
ThePerfectHacker
$10^{2y}=25$
Then,
$\left( 10^{2y} \right)^{1/2}=25^{1/2}=\sqrt{25}=5$
Thus,
$10^{2y\cdot 1/2}=10^y=5$
Raise both sides to (-1) thus,
$10^{-y}=5^{-1}=\frac{1}{5}$
• December 19th 2006, 07:22 PM
ThePerfectHacker
$(\log x)(\log_5 10)=3$
Divide both sides by the base 5 logarithm,
$\log x=\frac{3}{\log_5 10}$.
Now there is a nice little rule that says,
$\log_a b=\frac{1}{\log_b a}$
Thus, you can really write,
$\log x=3\log_{10} 5=3\log 5=\log 5^3$
Thus,
$x=5^3=125$
• December 19th 2006, 07:24 PM
ThePerfectHacker
$\log \frac{1}{2}+\log \frac{2}{3}+...+\log \frac{9}{10}$
Use the rule that the logarithm of a product is the sum of the logarithms,
$\log \left( \frac{1}{2}\cdot \frac{2}{3}\cdot ... \cdot \frac{9}{10} \right)$
Each one cancels,
$\log \frac{1}{10}=-1$
---
$x=\log 2$
Use the rule I taught you and take reciprocals,
$\frac{1}{x}=\log_2 10=\log_2 (2\cdot 5)=\log_2 2+\log_2 5$
Thus,
$\frac{1}{x}=1+\log_2 5$
Thus,
$\log_2 5=\frac{1}{x}-1$
• December 19th 2006, 07:36 PM
ThePerfectHacker
$2\log x=\log 2x$
Use the sum rule,
$2\log x=\log 2+\log x$
$\log x=\log 2$
$x=2$
---
$\log_3 (x+9)=2\log_5 (x-3)$
$\log_3 (x+9)=\log_5 (x-3)^2$
Change base,
$\log_5 (x+9)/\log_5 3=\log_5 (x-3)^2$
$\log_5 (x+9)=\log_5 (x-3)^2\cdot \log _5 3$
Thus,
$x+9=(x^2-6x+9)^{\log_5 3}$
I do not think this is solvable.
• December 19th 2006, 09:04 PM
Soroban
Hello, SportfreundeKeaneKent!

Quote:

If $x = \log_{10}2$, then express $\log_25$ in terms of $x$

We have: . $\log_{10}2\:=\:x$

Take reciprocals: . $\frac{1}{\log_{10}2} \:=\:\frac{1}{x}$ [1]

The left side is: . $\frac{1}{\log_{10}2} \:=\:\log_2(10) \:=\:\log_2(5\cdot2) \:=\:\log_2(5) + \log_2(2) \:=\:\log_2(5) + 1$

Then equation [1] becomes: . $\log_2(5) + 1 \;=\;\frac{1}{x}$

Therefore: . $\log_2(5)\;=\;\frac{1}{x} - 1\;=\;\frac{1-x}{x}$

• December 20th 2006, 07:17 AM
earboth
Quote:

Originally Posted by SportfreundeKeaneKent
...The problem that I'm having is that the base is different in some of them which is throwing me off.

Hello Sportsfreund,

your calculator "knows" the logarithms to the base 10 (that's the log-button) and the logarithms to the bease e (that's th ln-button).

You can transform the different logarithms by using the log or the ln-function of your calculator:

$\log_b{a}=\frac{\ln(a)}{\ln(b)}=\frac{\log(a)}{\lo g(b)}$ .

Quote:

Originally Posted by SportfreundeKeaneKent
By the way, the first question has a little typo. The right side of the equation should say 2log9(x-3) instead of 2log5(x-3)

http://img133.imageshack.us/img133/2...0132021kf7.png
...

According to your correction you'll solve:

$\log_3(x+9)=2\log_9(x-3)\ ,\ x>3$

I use the above mentioned formula:

$\frac{\ln(x+9)}{\ln(3)}=\frac{2\ln(x-3)}{\ln(9)}=\frac{2\ln(x-3)}{2\ln(3)}$

Multiply by ln(3):

$\ln(x+9)=\ln(x-3)$. This equation has no real solution.

I assume that there is maybe another typo.

EB
• December 21st 2006, 08:41 AM
SportfreundeKeaneKent
Alright, here's the answer to that typo question, by the way, you do change the exponent to 9:
logbase3(x+9)=logbase3(x-3)^2/log9base3
2logbase3(x+9)^2 - logbase3(x-3)^2=0
logbase3(x+9)^2/(x-3)^2=0
x^2-6x+9=x^2+18x+81
24x=-72
x=-3

I know that that's hard to follow but I think it's right.
• December 21st 2006, 11:48 AM
earboth
Quote:

Originally Posted by SportfreundeKeaneKent
Alright, here's the answer to that typo question, by the way, you do change the exponent to 9:
logbase3(x+9)=logbase3(x-3)^2/log9base3
2logbase3(x+9)^2 - logbase3(x-3)^2=0
logbase3(x+9)^2/(x-3)^2=0
x^2-6x+9=x^2+18x+81
24x=-72
x=-3

I know that that's hard to follow but I think it's right.

hello Sportsfreund,

it isn't so hard to follow your calculations. You calculated correctly. Nevertheless x = -3 is not a solution of your equation:

I copy the line where I can show you why:

"logbase3(x+9)=logbase3(x-3)^2/log9base3"

Now plug in x = -3. You'll get

logbase3(-3+9)=logbase3(-3-3)^2/log9base3

which is the same as

logbase3(6)=logbase3(-6)^2/log9base3

And it is impossible to calculate the logarithm of a negative number. Thus x = -3 is not a solution of your equation.

So sorry to disappoint you.

EB