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

17 Answers

Write as a single logarithm

also, I know the answer, I just don't know how to get it

the answer should be log_a(bc^4)

also, I know the answer, I just don't know how to get it

the answer should be log_a(bc^4)

# Answers

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

*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.

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.

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