# Basic Logirithms & Basic expanding the brackets

• Nov 6th 2009, 09:02 PM
student0451
Basic Logirithms & Basic expanding the brackets
Hello everyone,

I need help with logarithms that are in fraction form.

such as
evaluate the following for x:

log ^ 2 x = 1/7

log ^ 2 x = 1/5

log ^ 2 x = 2/7

I also need to be sure if expanding this equation for e.g is correct?

expand the following:

(a + b) (a + b + c)

a^2 + ab + ac + ba + b^2 + bc

& do I plus the powers or times them in the brackets?

expand the following:

(a^2 + b^2) (a^3 + b^3 + c^2)

a^5 + a^2b^5 + a^2c^2 + b^2a^3 + b ^ 5 + b ^2c^ 2
• Nov 6th 2009, 09:19 PM
Bacterius
Quote:

evaluate the following for x
You must mean "solve the following for x" ? If yes, then :

$\displaystyle log(2x) = \frac{\ 1}{7}$
$\displaystyle 2x = 10^{\frac{\ 1}{7}}$
$\displaystyle x = \frac{\ 10^{\frac{\ 1}{7}}}{2}$

$\displaystyle (a + b)(a + b + c)$
$\displaystyle a(a + b + c) + b(a + b + c)$
$\displaystyle a^2 + ab + ac + ba + b^2 + bc$
$\displaystyle a^2 + b^2 + 2ab + ac + bc$

But maybe I don't understand what you wrote, try using the LaTeX editor ?
• Nov 6th 2009, 09:19 PM
VonNemo19
Quote:

Originally Posted by student0451
Hello everyone,

I need help with logarithms that are in fraction form.

such as
evaluate the following for x:

log ^ 2 x = 1/7

log ^ 2 x = 1/5

log ^ 2 x = 2/7

I also need to be sure if expanding this equation for e.g is correct?

expand the following:

(a + b) (a + b + c)

a^2 + ab + ac + ba + b^2 + bc

& do I plus the powers or times them in the brackets?

expand the following:

(a^2 + b^2) (a^3 + b^3 + c^2)

a^5 + a^2b^5 + a^2c^2 + b^2a^3 + b ^ 5 + b ^2c^ 2

1. Do you mean

$\displaystyle \log_2x=\frac{1}{7}$ or $\displaystyle \log^2x=\frac{1}{7}$ ?

2. Collect your like terms
• Nov 6th 2009, 09:21 PM
student0451
i am so sorry about my typing,

http://www.mathhelpforum.com/math-he...6f55baf4-1.gif

it is the base of 2

log 2 x = 1/7

so 2 is the base of the log

and thank you for helping me.
• Nov 6th 2009, 09:24 PM
Bacterius
Then it becomes (I think) :

$\displaystyle log_2(x) = \frac{\ 1}{7}$
$\displaystyle x = 2^{\frac{\ 1}{7}}$

When writing indicators, use the underscore char, it is much more appropriate : log_2(x).
• Nov 6th 2009, 09:24 PM
VonNemo19
Quote:

Originally Posted by student0451
i am so sorry about my typing,

http://www.mathhelpforum.com/math-he...6f55baf4-1.gif

it is the base of 2

log 2 x = 1/7

so 2 is the base of the log

and thank you for helping me.

$\displaystyle \log_2x=\frac{1}{7}\Rightarrow{x}=2^{1/7}=\sqrt[7]{2}$
• Nov 6th 2009, 09:28 PM
student0451
yep

and thank you, vondera19 & Bacterius.

I knew there was something wrong between the way I expanded brackets.