Finding log something, in terms of A and B.

In summary, the conversation discusses the solution to a bonus question in a geometric quiz. The question involves finding the value of logb397 in terms of A and B, where Logb2=A and Logb49=B. The conversation includes a calculation using the geometric formula arn-1 and concludes with a question about the accuracy of the solution.
  • #1
Suy
101
0

Homework Statement



If Logb2=A and Logb49=B, what is logb397, 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
bA=2 and bB=49
b=21/A b=491/B
491/B=21/A --> log(49)/log(2)=B/A

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

log(2)(B/A)n-1=logb397
am i right?
 
Last edited:
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  • #2
sorry wrong post...
 

Related to Finding log something, in terms of A and B.

1. What is the formula for finding log something in terms of A and B?

The formula for finding log something in terms of A and B is: logA(B).

2. How do I solve for A or B in the equation log something = A * log(B)?

To solve for A or B in the equation log something = A * log(B), you can use the property of logarithms which states that log(Ax) = x * log(A). Therefore, to solve for A, divide both sides by log(B) and for B, divide both sides by log(A).

3. What is the purpose of using logarithms in equations?

Logarithms are used to simplify complex equations and to solve for unknown variables. They also help to make it easier to compare values that vary over a wide range.

4. Can I find log something without knowing the values of A and B?

No, in order to find log something in terms of A and B, you need to know the values of both A and B. However, if you know the values of log something and one of the variables (either A or B), you can solve for the other variable using algebraic methods.

5. What is the relationship between logarithms and exponents?

Logarithms and exponents are inverse operations of each other. This means that if you have an equation in the form of logA(B) = C, you can rewrite it in exponential form as AC = B. They essentially undo each other, making it easier to solve for unknown variables.

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