Question

Let \[\log_{b}A=1 ; \log_{b}C=3 ; \log_{b}D=4 \] . What is the value of \[\log_{b}(\frac{ A^5C^2 }{ D^6 } )\] ?

A. 0
B. 26
C. -13
D. There isn't enough information to answer the question.

Answer 1

@Hero do you think you could please help me? :)

Answer 2

Okay so basically what you do is use rules of logs to rewrite it as:

\[\log_b(A^5) + \log_b(C^2) - \log(D^6)\]

Then as:

\[5\log_bA + 2\log_bC -6\log_bD\]

Then substitute the appropriate values an finish solving.

Answer 3

I keep getting logb (9/4096) but that doesn't seem right... what am I doing wrong?

Answer 4

Then you end up with

5(1) + 2(3) - 6(4) = 5 + 6 - 24
= 11 - 24
= -13

Answer 5

Oh I was substituting the wrong numbers in the wrong places! I see what you did now!

Answer 6

I'm not an expert at these, but I think that should be correct.

Answer 7

It was! thank you SO much :) do you know how to rewrite an expression of "log's" into a single logarithm ?

Answer 8

I've done it before

Answer 9

Rewrite the following expression as a single logarithm: 2log(x+3)+3log(x-7)-5log(x-2)-log(x^2)

A. \[\log(\frac{ (x+3)^2(x-7)^3 }{ x^2(x-2)^5 })\]

B. \[\log(\frac{ x^2(x+3)^2(x-7)^3 }{ (x-2)^5 })\]

C. \[\log(\frac{ (x+3)^2(x-2)^5 }{ (x-7)^3x^2 })\]

D. \[\log(\frac{ (x+3)^2(x-7)^3 }{ (x-2)^-5x^-2 })\]

Answer 10

It's not that difficult at all

Answer 11

what do i do? ive never done this

Answer 12

First convert 2log(x+3)+3log(x-7)-5log(x-2)-log(x^2) to

\[\log(x + 3)^2 + \log(x - 7)^3 - \log(x - 2)^5 - \log(x^2)\]

Then convert that to
\[\log((x + 3)^2(x - 7)^3) - \log\left(\frac{(x - 2)^5}{x^2}\right)\]

Which then becomes:
\[\log(a) - \log(b) = \log\left(\frac{a}{b}\right)\]

Where a = \((x + 3)^2(x - 7)^3\) and b = \(\dfrac{(x - 2)^5}{x^2}\)

Answer 13

So perform this division:

\[a \div b = (x + 3)^2(x - 7)^3 \div \dfrac{(x - 2)^5}{x^2}\]

Answer 14

Afterwards, put the log in front.

Answer 15

I got A. Is that correct?

Answer 16

Can you show your work please?

Answer 17

well I thought that because you cant divide by an x you would just leave all of the functions however they were in the equation and since you cant fivide by x just add it to (x-2)^5 since it was a part of that function which would make it (x-2)^5+x^2

Answer 18

right?

Answer 19

So you didn't divide by hand. There's only one way to divide by a fraction.


\[a \div \frac{b}{c} = a \times \frac{c}{b}\]

Answer 20

I'm not in a place where I have writing utensils available so no, but I see what you did there, you divided using the reciprical/inverse right?

Answer 21

What do you mean you're not in a place where you have writing utensils. You don't do math without such things.