1. ## logarithms?

I have no idea how to do these problems, or even how to write them correctly in text format. heh.

Okay so,

1. It says:
If log(base 8)3 = P and log(base)35 = Q, express log(base 10)5 in terms of P and Q.

I hope that makes sense.

Four TA's couldnt figure this one out.

I THINK my teacher said the answer is [3-3I]? (I dont think thats an imaginary i) because he wrote it in a capital I. Maybe he messed it up, I dont know, regardless, I just need all the steps of this problem and understand how to do it.

2. says:
Given that log(base)4n 40√3 = log(base)3n45, find n^3. (exponent)

yeah I have no idea how to do that. I'm not going to lie. heh. It was never explained nor is it in our book?

So if anyone could help that would be greatly appreicated. :[

I have no idea how to do these problems, or even how to write them correctly in text format. heh.

Okay so,

1. It says:
If log(base 8)3 = P and log(base)35 = Q, express log(base 10)5 in terms of P and Q.
i don't get what was so hard about this problem. i'm not a smart guy, so that's saying something.
when asked to express something in terms of something else, it means you want to end up with something = something else. so the question asked to express log(base 10) 5 in terms of P and Q, that means we want log(base 10)5 = somefunction of P and Q

i will write log[a]b to mean log to the base a of b.

we have log[8]3 = P
=> 8^P = 3

we have log[3]5 = Q
=> 3^Q = 5

Now since 3 = 8^P we can replace the 3 in the last equation with 8^P, so we have (8^P)^Q = 8^PQ = 5
now take log[10] of both sides, we get

log[10]8^PQ = log[10]5
=> log[10]5 = PQlog[10]8

OR

log[8]3 = P log[3]5 = Q
apply the change of base formula and change both to base 10, we get

log[10]3/log[10]8 = P and log[10]5/log[10]3 = Q

so we have log[10]3 = Plog[10]8 and log[10]3 = log[10]5/Q
equating both these equations we get:

Plog[10]8 = log[10]5/Q ........now multiply both sides by Q
=> PQlog[10]8 = log[10]5
=> log[10]5 = PQlog[10]8

that wasn't that hard, you scared me with the TA thing

did you understand the steps. i did it by two methods, hopefully you get one of them. for the first i used the definition of a logarithm, for the second, i used the change of base formula

3. Theres a lot more steps to it though, my professor did a similar problem on the chalk board, which no one else in the class understood either, and he came up with an answer with an imaginary number. The problem took up the whole board. but what about the second one?

Theres a lot more steps to it though, my professor did a similar problem on the chalk board, which no one else in the class understood either, and he came up with an answer with an imaginary number. The problem took up the whole board. but what about the second one?
i dont really see any other way to do the problem. as for your professor's answer, it has to be wrong, since there is no P or Q in 3 - 3I, obviously he was solving a different type of problem. perhaps one similar to the second one--which at the moment, i haven't figured out what to do with

I have a solution to #1, but it's very convoluted . . .

If log83 = P and log35 = Q, express log105 in terms of P and Q.

log
83 = P . . 8^P = 3 . . (2³)^P = 3 . . 2^(3P) = 3

. . Raise both sides to the power Q:
. . . . [2^(3P)]^Q = 3^Q . . 2^(3PQ) = 3^Q .[1]

log
35 = Q . . 3^Q = 5 .[2]

Equate [2] and [1]: .5 .= .2^{3PQ)

Take logs (base 10): .log
10(5) .= .log10[2^(3PQ)] .= .(3PQ)log10(2)

Therefore: . log
10(5) .= .3·P·Q·log10(2)

Edit: Wow, Jhevon, I like your solution!

6. Originally Posted by Soroban

I have a solution to #1, but it's very convoluted . . .

log
83 = P . . 8^P = 3 . . (2³)^P = 3 . . 2^(3P) = 3

. . Raise both sides to the power Q:
. . . . [2^(3P)]^Q = 3^Q . . 2^(3PQ) = 3^Q .[1]

log
35 = Q . . 3^Q = 5 .[2]

Equate [2] and [1]: .5 .= .2^{3PQ)

Take logs (base 10): .log
10(5) .= .log10[2^(3PQ)] .= .(3PQ)log10(2)

Therefore: . log
10(5) .= .3·P·Q·log10(2)

i got the same answer for #1, #2 is the problem now, do you see what to do with that Soroban?

I think I got #2 . . .

2. Given that: . log4n(40√3) .= .log3n(45), .find n³.

Let .log
4n(40√3) .= .log3n(45) .= .P

Then we have: .log
4n(40√3) = P . . (4n)^P = 40/3 .[1]
. . . . . . . and: . .log
3n(45) = P . . . (3n)^P = 45 - -[2]

. . . . . . . . . . . . . .4^P·n^P . . . 40√3
Divide [1] by [2]: . ----------- .= .------ . . (4/3)^P .= .(4/3)^{3/2}
. . . . . . . . . . . . . .3^P·n^P. . . . .45

. . Hence: .p = 3/2

Substitute into [2]: .(3n)^{3/2} .= .45

Raise both sides to the 2/3 power: .3n .= .(45)^{2/3} .= .3^{4/3}·5^{2/3}

Divide by 3: .n .= .3^{1/3}·5^{2/3}

Cube both sides: . .= .3·5² . . n³ = 75

8. Originally Posted by Soroban

I think I got #2 . . .

Let .log
4n(40√3) .= .log3n(45) .= .P

Then we have: .log
4n(40√3) = P . . (4n)^P = 40/3 .[1]
. . . . . . . and: . .log
3n(45) = P . . . (3n)^P = 45 - -[2]

. . . . . . . . . . . . . .4^P·n^P . . . 40√3
Divide [1] by [2]: . ----------- .= .------ . . (4/3)^P .= .(4/3)^{3/2}
. . . . . . . . . . . . . .3^P·n^P. . . . .45

. . Hence: .p = 3/2

Substitute into [2]: .(3n)^{3/2} .= .45

Raise both sides to the 2/3 power: .3n .= .(45)^{2/3} .= .3^{4/3}·5^{2/3}

Divide by 3: .n .= .3^{1/3}·5^{2/3}

Cube both sides: . .= .3·5² . . n³ = 75

Brilliant!

I did the same as you did up to around the third line, but then i tried to do something weird and it didn't work out.

Bravo!

9. Okay thanks very much, I hope you're right.

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### give that log4n40

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