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Old November 6th, 2009, 10:02 PM
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Default 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
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  #2  
Old November 6th, 2009, 10:19 PM
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Quote:
evaluate the following for x
You must mean "solve the following for x" ? If yes, then :

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

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

But maybe I don't understand what you wrote, try using the LaTeX editor ?
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Old November 6th, 2009, 10:19 PM
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Quote:
Originally Posted by student0451 View Post
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

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


2. Collect your like terms
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Old November 6th, 2009, 10:21 PM
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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.
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  #5  
Old November 6th, 2009, 10:24 PM
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Default Reply

Then it becomes (I think) :

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

When writing indicators, use the underscore char, it is much more appropriate : log_2(x).
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Old November 6th, 2009, 10:24 PM
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Quote:
Originally Posted by student0451 View Post
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.
\log_2x=\frac{1}{7}\Rightarrow{x}=2^{1/7}=\sqrt[7]{2}
__________________
"Then thou carriedst thine ashes into the mountains:
wilt thou now carry thy fire into the valleys?"

Thus Spake Zarathustra
Friedrich Nietzsche

If you would like to know how the quadratic formula was derived, visit my post entitled:
Deriving the Quadratic Formula.


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Old November 6th, 2009, 10:28 PM
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yep

and thank you, vondera19 & Bacterius.

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