# The Expression Log_A(B) + 4Log_A(Ac) – 4, Where A, B, C > 1, Written As A Single Logarithm, Is

habibineedsome | 07:46 Thu 17th Jun 2021 | Jobs & Education
Write as a single logarithm
also, I know the answer, I just don't know how to get it

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One of the keys to solving maths problems is to stop obsessing over "know[ing] how to get" the answer, and start focusing instead on knowing how to *try*. Here, we are working with logarithms, and the tools at our disposal are the "laws of logs". In no particular order, they are (bases omitted except when it needs to be made explicit, and all numbers strictly...
08:06 Thu 17th Jun 2021
One of the keys to solving maths problems is to stop obsessing over "know[ing] how to get" the answer, and start focusing instead on knowing how to *try*. Here, we are working with logarithms, and the tools at our disposal are the "laws of logs". In no particular order, they are (bases omitted except when it needs to be made explicit, and all numbers strictly positive):

(1): log(xy) = log(x) + log(y) ;
(2): log(x/y) = log(x) - log(y) ;
(3): log(x^y) = y log(x) ;
(4): log(1) = 0 ;
(5): log_x(x) = 1 for all x.

The only thing we can do when presented with a problem is to seek to apply these, in some order. I don't know what the right order is, so I'll just try. For example, first you might decide to apply law (3) above, and write:

4log_A(AC) = log_A((AC)^4)

Then you can see that log_A(B) + log_A((AC)^4) can be combined using law (1), giving

log_A(B) + log_A((AC)^4) = log_A(B*(AC)^4)

And at this point you might feel that you are stuck, because there's still that irritating "-4" at the end. I'll point out a trick in a separate post to deal with this, because it's far from obvious, but what I'm wanting to emphasise is that I have *tried* something, nothing magic, just following the known laws and seeing where they take me. If the answer is (or seems to be) nowhere, then let's try something else.

* * * * * *
Apply law (1) to 4Log_A (AC), to write:

4log_A(AC) = 4(Log_A(A) + log_A(C))

Use ordinary algebra to expand this:

4(log_A(A) + log_A(C)) = 4 log_A(A) + 4 log_A(C)

At this point, you should recognise that we can apply law (5) above to replace log_A(A) = 1, so that 4 log_A(A) = 4. Put this into the rest of the expression, and get:

log_A(B) + 4 log_A(AC) - 4 = log_A(B) + 4 + 4 log_A(C) - 4

the +4 and -4 cancel, and we are left with *only terms containing logarithms! The rest, we can do by applying the laws a couple more times.

I can't stress this enough: I didn't know every step, in every order, on the way to this answer when I set out to solve this problem. I just had some tools, and I knew I had to apply them, and let the maths itself take care of the rest.
The "trick" I was referring to in my first attempt is that you can apply law (5) backwards. There is nothing to stop you, ever, from writing and using:

1 = log_x(x)

It follows that every number, ever, can be written as

N = N*1 = N*log_x(x)

Following on from what I did above, I could therefore write:

log_A(B) + 4log_A((AC)) - 4
= log_A(B*(AC)^4) - 4
= log_A(B*(AC)^4) - 4 log_A(A)

Applying law (3), we can write this again as
= log_A(B*(AC)^4) - log_A(A^4)

And I'll leave the rest for you to try.
Amazing & beautiful jim! to the Greeks, mathematics was considered to be one of the arts.
Well done jim. If the OP is a serious student he will recognise the teaching and effort you put into your answer; if he just wants his homework doing for him you'll probably hear no more.
Fantastic effort, jim. Love it.
Question Author
Yoo @jim360 THANKS, BRO! I followed what you had said, and managed to have it click in my head. I am able to get the answer to the question as well as the answer to similar questions due to your advice. Once again I thank you!
Feel free to ask any other questions that you wish in other threads, although it is worth noting that the maths questions don't always get much attention -- or much quick attention, at least. At the risk of undercutting this site, I can highly recommend places like r/askmath or r/MathHelp on reddit.com, which tend to be more active and quicker at providing help and advice on a range of maths questions.