
Subject: logarithms
Can you please answer those three question
 inverse or the exponential function.
logx 81 = 4,
x expo4 = 81,
x = +or 4square root 81,
x = +or 3
s.s = {3}
Why is 3 not acceptable.
 log3 ( x square  1) = 1,
3expo 1 = x square  1
x square = 4
x = +or square root 4
x = +or 2
s.s = {2,2}
Why is  2 acceptable here.
 Express as a single logarithm.
3.loga r  0.5.loga 5 + 0.33.loga t
Does the dot after 3&.5&.33 means to multiply, if so how do I proceed. If not what does it means.
Thank you
Pierre.
Hi Pierre,
 If x is greater than zero then the logarithm with base x is defined by
log_{x} a = y if and only if x^{y} = a
The reason that log_{x} is not defined for negative x is the expression x^{y}. If x is negative then x^{1/2}, x^{1/4}, x^{3/4}, x^{1/8}... are not real numbers.
 When you solve an equation, as you have done in this problem, you should, if possible, verify that the "solutions" you get are actually solutions to the original problem. For this problem
log_{3} ( x^{2}  1) = 1
substituting x = 2 or x = 2 on the left side gives
log_{3}3
which is 1. Thus x = 2 and x = 2 are both solutions.
 Yes I think the dots you specify are multiplication. I think the expression is (writing * for multiplication)
3*log_{a} r  (1/2)*log_{a} 5 + (1/3)*log_{a} t
The crutial fact here is the rule of logarithms that says
log_{a} x^{y} = y*log_{a} x
Hence
3*log_{a} r  (1/2)*log_{a} 5 + (1/3)*log_{a} t
= log_{a} r^{3}  log_{a} 5^{1/2} + log_{a} t^{1/3}
Now use the other two rules of logarithms to express this as a single logarithm.
Cheers,
Harley
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