logarithms

i) log ₇49 = (log ₃49) / (log ₃7) = [(log ₁₀49) / (log ₁₀3) ] / (log ₃7) = log ₇(7 x 7)= log ₇7 + log ₇7 = 2, since log _xx=1

log ₃81 = (log ₁₀81) / (log ₁₀3) = log ₃(3 x 3 x 3 x 3) = log ₃3 + log ₃3 + log ₃3 + log ₃3 = 4

log ₃1 = (log ₁₀1) / (log ₁₀3) = log ₃(3/3) = log ₃3 - log ₃3 = 0


ii) log ₇49^X / 343^y == > log ₇(49^X) = X*log ₇49 = X * 2 = 2X | (log ₇49)^X = 2^X <=== So which one is it?

dee:

simplify each of the following expressions

i) log[size=50]7[/size]49 + log[size=50]3[/size]81 + log[size=50]3[/size]1

ii) log[size=50]7[/size]49^x / 343^y

help please ive an exam tomorrow

Another way to do part 1 is to understand that
A LOGARITHM IS AN EXPONENT (POWER)

7 ²=49 therefore log ₇49=2
3 ⁴=81 therefore log ₃81=4
3 ⁰=1 therefore log ₃1=0

As Zamarronics said, the second one is ambiguous, is it [log[size=50]7[/size]49 ] ^x/ 343^y or log[size=50]7[/size] [49 ^x ] / 343^y

It would be good for you to work through this web page

http://www.mathsisfun.com/algebra/logarithms.html