Finding log something, in terms of A and B.

[Suy]

Homework Statement

If Log_b2=A and Log_b49=B, what is log_b397, in terms of A and B.

This is one of the bonus question in my geometric quiz and i don't remember if the number is 397. I wonder if anyone get this?

Homework Equations

The Attempt at a Solution

here is what i did
b^A=2 and b^B=49
b=2^1/A b=49^1/B
49^1/B=2^1/A --> log(49)/log(2)=B/A

because this is geometric quiz, so i assume this one have a geometric
so i use ar^n-1
r : log(49)/log(2)=B/A
a : Log_b2

Log_b2(B/A)^n-1=log_b397
am i right?
ty!

 

[Mark44]

Homework Statement

If Log_b2=A and Log_b49=B, what is log_b397, in terms of A and B.

This is one of the bonus question in my geometric quiz and i don't remember if the number is 397. I wonder if anyone get this?

I'm guessing that the number is 392, not 397. 392 = 8*49 = 2³*49.

Homework Equations

The Attempt at a Solution

here is what i did
b^A=2 and b^B=49
b=2^1/A b=49^1/B
49^1/B=2^1/A --> log(49)/log(2)=B/A

because this is geometric quiz, so i assume this one have a geometric
so i use ar^n-1

This makes no sense whatever. From your presentation of the problem, it has nothing to do with a geometric sequence, or anything else having to do with geometry. This problem is strictly concerned with the properties of logarithms.

r : log(49)/log(2)=B/A
a : Log_b2

Log_b2(B/A)^n-1=log_b397
am i right?
ty!

I'm assuming that this is the actual problem description:

If Log_b2=A and Log_b49=B, what is log_b392, in terms of A and B.​

log_b 392 = log_b (8 * 49) = log_b (2³ * 49)

Now, use the properties of logs on the last expression above to get quantities that are in terms of A and B.

 

[Suy]

but i remember it is a odd number

 

[Mark44]

Well, if you can't remember exactly what the problem is, I can't help you.

 

[Suy]

but what happen when it is a odd number? like 397, is it possible to solve it?

 

[Mark44]

If you don't know what the number is, you can't work the problem - that's what happens.

 

[Suy]

ok, thx