Logarithm base transformation I am trying to solve a problem which, I think, revolves around base transformation of logarithms. It goes like this: log_5 log_6 ((6x−1)/(x+1))<log_(1/5)log_(1/6)((x+1)/(6x−1)) I tried transforming "first" logarithms to base 5, yielding log_6((6x-1)/(x+1))<1/(log_(1/6)*(x+1)/(6x-1))) (if I am right, of course..) Further transformation to base 6 leaves me helpless with :

shaunistayb1

shaunistayb1

Answered question

2022-09-25

Logarithm base transformation
I am trying to solve a problem which, I think, revolves around base transformation of logarithms. It goes like this:
log 5 log 6 6 x 1 x + 1 < log 1 5 log 1 6 x + 1 6 x 1
I tried transforming "first" logarithms to base 5, yielding
log 6 6 x 1 x + 1 < 1 log 1 6 x + 1 6 x 1
(if I am right, of course..) Further transformation to base 6 leaves me helpless with :
log 6 6 x 1 x + 1 < 1 log 6 x + 1 6 x 1
Thanks.

Answer & Explanation

Dillon Levy

Dillon Levy

Beginner2022-09-26Added 12 answers

First of all, the existence conditions: the argument of a logarithm must always be > 0. So we must impose
log 6 6 x 1 x + 1 > 0 , log 1 6 x + 1 6 x 1 > 0
But observe that log 1 6 x + 1 6 x 1 = log 6 6 x 1 x + 1 (from the changing base formula... see later!); hence we can work only on the first.
The same holds for this one: the arguments needs to be > 0... but we want also this logarithm > 0 so the conditions on the argument is simply
6 x 1 x + 1 > 1 , i.e. x > 5 / 2
Moreover x must be different from 1 (the denominator has to be different from zero), but this is included in x > 5 / 2
Let's now go to the computations: you approached to
log 6 6 x 1 x + 1 < 1 log 6 x + 1 6 x 1
using the change base formula for logarthims: log a x = log b x log b a . And you're right.
Now note that
1 log 6 x + 1 6 x 1 = 1 log 6 x + 1 6 x 1 = 1 log 6 6 x 1 x + 1
where the last equality follows from the basic properties of logarithms, i.e. log a x = log a 1 x
So our inequality is now turned in the following one:
log 6 6 x 1 x + 1 < 1 log 6 6 x 1 x + 1
Now simply multiply every side for log 6 6 x 1 x + 1 , so we have:
( log 6 6 x 1 x + 1 ) 2 < 1
i.e.
1 < log 6 6 x 1 x + 1 < 1
Then taking the power of 6 of all sides we came to
1 6 < 6 x 1 x + 1 < 6
that leads to x > 1 / 5. But the existence conditions impose that x > 2 / 5
Hence the solution is x > 2 / 5
Darius Miles

Darius Miles

Beginner2022-09-27Added 3 answers

You need to use the following two properties of the logarithm: log b y = log 1 / b y and log b y = log b 1 / y
Using these properties the right hand side becomes: log 1 / 5 log 1 / 6 x + 1 6 x 1 = log 1 / 5 log 6 6 x 1 x + 1 = log 5 log 6 6 x 1 x + 1
So your inequality now becomes log 5 log 6 6 x 1 x + 1 < 0 which is quite trivial to solve.

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