No. Insert the word "minus" in place of the word "plus", and you'll have a true statement.
Not quite. The log(x/y) = log(x) - log(y) In words, this reads "The log of a quotient is the difference of the log of the numerator and the log of the denominator."
log4+log3=log(4x3)=log12
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
1.268293446
You use the identities: log(ab) = b log(a), and log(ab) = log a + log b. In this case, 5 log42 + 7 log4x + 4 log4y = log432 + log4x7 + log4y4 = log4 (32x7y4).
For a quotient x/y , then its log is logx - log y . NOT log(x/y)
Not quite. The log(x/y) = log(x) - log(y) In words, this reads "The log of a quotient is the difference of the log of the numerator and the log of the denominator."
False When logs are taken, division becomes subtraction, so the log of a quotient is the log of the numerator minus the log of the denominator.
True. For example: 4 X 104/2 X 108 = 2 X 10-4
It has no specific name. For example f(x) = sin(x)/log(x) where x not equal to 1
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
log4+log3=log(4x3)=log12
That is the same as log xy.
you can't
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
1.268293446
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4