Problems with logaritm

• Oct 21st 2008, 09:23 PM
blubla
Problems with logaritm
Hello everyone! I have a few problem regarding Log.
First of all, it is a homework about Logarithm. I provide an image of my homework here:
http://img443.imageshack.us/img443/8...thmsbj4.th.jpg
(The ones squared with red are the ones I have problem with solving)
The main problem I have with Logs are the exponents. I know how to convert from logarithm form to exponent form and vice versa. But when solving for a variable exponent, it becomes difficult.
This is what I have done:
http://img75.imageshack.us/img75/9156/log1kf8.th.jpg
In problem 57.
I can solve a.
But b, c, and d are different.
In b. when solving on exponential form, how can I make a base with a variable exponent equal to a fraction (or decimal)?
In c. when solving on exponential form, how can I make a base with a variable exponent equal to a root?
The same goes with d.
In problem 58.
I could solve a. But it seems that b, c, and d, are solved differently. All of them contain negative exponents. How can I make a base with an exponent equal to another positive number if the exponent is negative?
Thanks in advance for the help!
• Oct 21st 2008, 10:17 PM
earboth
Quote:

Originally Posted by blubla
Hello everyone! I have a few problem regarding Log.
First of all, it is a homework about Logarithm. I provide an image of my homework here:
...
(The ones squared with red are the ones I have problem with solving)
The main problem I have with Logs are the exponents. I know how to convert from logarithm form to exponent form and vice versa. But when solving for a variable exponent, it becomes difficult.
This is what I have done:
...

to b):

$\displaystyle \log_2\left(\dfrac1{16}\right)=\log_2\left(\dfrac1 {2^4}\right)=\log_2\left(2^{-4}\right)=-4$

The other two problems have to be done similarly.

to c):

$\displaystyle \log_2(\sqrt{8})=\log_2(\sqrt{2^3})=\log_2((2^3)^{ \frac12})=\log_2(2^{\frac32})=\dfrac32$

The other two problems have to be done similarly.

to d):

$\displaystyle \log_{\frac12}(32)=\log_{\frac12}(2^5)=\log_{\frac 12}(((2^{-1})^{-1})^5)=\log_{\frac12}\left(\left(\dfrac12\right)^{-5}\right) = -5$

The other two problems have to be done similarly.
• Oct 21st 2008, 10:40 PM
earboth
Quote:

Originally Posted by blubla
Hello everyone! I have a few problem regarding Log.
...
In problem 57.
I can solve a.
But b, c, and d are different.
In b. when solving on exponential form, how can I make a base with a variable exponent equal to a fraction (or decimal)?
In c. when solving on exponential form, how can I make a base with a variable exponent equal to a root?
The same goes with d.
In problem 58.
I could solve a. But it seems that b, c, and d, are solved differently. All of them contain negative exponents. How can I make a base with an exponent equal to another positive number if the exponent is negative?
Thanks in advance for the help!

If you have the equation

$\displaystyle x^a = c$ ....... which you want to solve for x you only have to take both sides to the power $\displaystyle \dfrac1a$

That means: $\displaystyle x^a = c~\implies~\left(x^a\right)^{\frac1a} = c^{\frac1a}~\implies~x^1=c^{\frac1a}$

to 58 b):

$\displaystyle x^{-3} = 2~\implies~\left(x^{-3}\right)^{-\frac13} = 2^{-\frac13}~\implies~x=\dfrac1{\sqrt[3]{2}}$

to 58 c):

$\displaystyle x^{-8} = \dfrac1{16} ~\implies x^{-8} = \dfrac1{2^4} \implies x^{-8} = 2^{-4} \implies$ $\displaystyle \left(x^{-8}\right)^{-\frac18} = \left(2^{-4}\right)^{-\frac18} \implies x=2^{\frac12} \implies x=\sqrt{2}$

to 58 d):

Keep in mind that 27 = 3³. Therefore the solution is $\displaystyle x = \dfrac13$
• Oct 23rd 2008, 12:05 AM
blubla
That helped me a lot. Thanks!(Nod)