# Trouble With Simple Logarithms :(

• Nov 14th 2008, 07:50 PM
jimzer
Trouble With Simple Logarithms :(
hey guys

everytime i solve this particular question i end up with x = 1/2

but it's wrong and the answer is actually 3 and i want to know where i went wrong, thanks a lot in advance
• Nov 14th 2008, 07:55 PM
mr fantastic
Quote:

Originally Posted by jimzer
hey guys

everytime i solve this particular question i end up with x = 1/2

but it's wrong and the answer is actually 3 and i want to know where i went wrong, thanks a lot in advance

Note that:

1. \$\displaystyle \log_2 8 = 3\$

2. \$\displaystyle \log_2 8^{1-x} = (1 - x) \log_2 8\$.

By the way, what you've posted isn't an equation ....What's the expression meant to be equal to ...?
• Nov 14th 2008, 07:57 PM
jimzer
the book asks to simplify without a calculator, and thanks a lot for that swift reply
• Nov 14th 2008, 07:59 PM
mr fantastic
Quote:

Originally Posted by jimzer
the book asks to simplify without a calculator, and thanks a lot for that swift reply

Do you see now why the answer is 3 ....?
• Nov 14th 2008, 08:01 PM
jimzer
call me silly but no :(

i see how log(2)8 = 3 cause 2^3 = 8

but there's the x's and i have no clue how to simplify or solve this question :/
• Nov 14th 2008, 08:35 PM
mr fantastic
Quote:

Originally Posted by jimzer
call me silly but no :(

i see how log(2)8 = 3 cause 2^3 = 8

but there's the x's and i have no clue how to simplify or solve this question :/

Look at what I posted. Your expression becomes x(3) + (1 - x)(3) = ....
• Nov 14th 2008, 09:14 PM
jimzer
.. i cant believe i didnt see that!!

THANKS SO MUCH AGAIN!!