I'm stuck

[MethMath11]

Progress so far =
Log(B) 12/ Log(A) 12

 

[JeffM]

The problem is shown through a picture that is not properly oriented and somewhat fuzzy. The problem given is in an unspecified language and appears to use a mathematical notation that differs somewhat from that used in the U.S.

You may have to wait for a detailed answer until a tutor comes along who is familiar with your language and notation.

Does [MATH]^Alog(S)[/MATH] mean the logarithm to the base A of S? If so, the standard notation for that in the U.S. is

[MATH]log_A(S) \text { or } log_A \ S[/MATH].

 

[MethMath11]

The problem is shown through a picture that is not properly oriented and somewhat fuzzy. The problem given is in an unspecified language and appears to use a mathematical notation that differs somewhat from that used in the U.S.

You may have to wait for a detailed answer until a tutor comes along who is familiar with your language and notation.

Does [MATH]^Alog(S)[/MATH] mean the logarithm to the base A of S? If so, the standard notation for that in the U.S. is

[MATH]log_A(S) \text { or } log_A \ S[/MATH].

Yes,

 

[JeffM]

Please try to write the problem in English. I do not know what it is asking. In fact, I do not even recognize the language in which it is written so I cannot hope to translate it. I'd like to help, but the poor picture and unknown language prevent that.

And does [MATH]log(x)[/MATH] mean [MATH]log_{10}(x)[/MATH] or [MATH]log_e(x).[/MATH]

 

[Dr.Peterson]

I believe that the problem is this (in American notation):

If [MATH]\log_A(5) = 0.888...[/MATH] and [MATH]\log_B(5) = 0.3232...[/MATH], then [MATH]\frac{\log_B(36) - \log_B(3)}{(\log_A(2) + \log_A(5))(\log_{10}(2) + \log_{10}(6))} = [/MATH]?​

What have you tried?

I have reasons to assume that your log means base ten; in order to make use of that, I would try expressing everything in terms of base 10. I haven't yet solved the problem, though.

 

[Dr.Peterson]

Progress so far =
Log(B) 12/ Log(A) 12

I didn't see (or understand) this piece before. Now that I have solved the problem, I see that it shows you have done much of the work. (It would have been very helpful if you had said more about how you got here.)

Now do what I suggested, and use the change-of-base formula to express both logs in terms of base ten. Then do the same with the given facts about logs of 5 to write expressions for the base-ten logs of A and B, and put those into the expression you got.

 

[HallsofIvy]

First, a little simplification. If x= 0.8888... then $\displaystyle x= \frac{8}{9}$.
If $\displaystyle log_A(5)= \frac{8}{9}$ then $\displaystyle A^{\frac{8}{9}}= 5$. To solve for A take the logarithm of both sides, (8/9)Log(A)= log(5) , using whichever base you prefer, common log or natural log.

Using the common logarithm, log(5)= 0.6990 (to four decimal places) so $\displaystyle log(A)= \frac{9}{8}(0.6990)= 0.7863$ and $\displaystyle A= 10^{0.7863}= 6.114$.

Using the natural logarithm, ln(5)=1.609 so $\displaystyle ln(A)= \frac{9}{8}(1.609)= 1.8106$ and $\displaystyle A= e^{1.8106}= 6.114$ again.

 

[Dr.Peterson]

Note that it will not be necessary to actually calculate values for log(A) or log(B); expressions involving log(5) and log(12) will be sufficient, as they will cancel out in the end. I would just apply the change of base formula four times, then do a little juggling.