Solving log(base5)36 using log2 and log3

  • Thread starter seiferseph
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In summary, when given the equations log2 = x and log3 = y, and asked to solve for log(base5)36 in terms of x and y, you can use the formula log(ab) = log(a) + log(b) to simplify the equation to log(base5)36 = 2x + 2y / log(base10)5. This can be further simplified to log 5 = 1-x.
  • #1
seiferseph
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if log2 = x and log3 = y, solve for log(base5)36 in terms of x and y.

can someone help me get started with this one? thanks.
 
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  • #2
Try to use the translation of base formula.
Or let log(base 5)10=z
Try to think of how to convert log5 in terms of x.
Notice that some special value you can get, such as log2=x , log3=y, log1=0, log 10=1,etc.
Then you can express it in term of x.
 
  • #3
What is the base in the original log, 10?

Assuming you mean log10(2)= x and log10(3)= y,

log5(10)= 1/log10(5).

5= 10/2 so log10(5)= log10(10/2)= log10(10)- log10(2)= 1- x.

log10(3) doesn't enter into it.
 
  • #4
i'll post a little bit of what I've done, the teacher said its simple, and in the last questions we've converted the bases for x and y to the one for the final, not the other way around. here's what I've done, not sure if its right.
 

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  • #5
sorry, its supposed to be log(base5)36 to solve for
 
  • #6
36 = 3*3*2*2

Now apply the fact that: log(ab) = log(a) + log(b)
 
  • #7
Zurtex said:
36 = 3*3*2*2

Now apply the fact that: log(ab) = log(a) + log(b)

so in the end i get

log(base5)36 = 2x + 2y / log(base10)5
can it be simplified further?
 
  • #8
Yes.
log 5=1-x
 
  • #9
primarygun said:
Yes.
log 5=1-x
You sure you have read the edits?
 
  • #10
You sure you have read the edits?
What's edited?
 
  • #11
primarygun said:
What's edited?
The original question.
 

FAQ: Solving log(base5)36 using log2 and log3

1. How do I solve log(base5)36 using log2 and log3?

To solve this problem, we can use the change of base formula, which states that log(basea)b = logc(b)/logc(a). In this case, we want to find log(base5)36, so we can use either log2 or log3 as our base. Let's use log2 for this example. We can rewrite the problem as log2(36)/log2(5). Using a calculator, we can find that log2(36) = 5.1699 and log2(5) = 2.3219. Therefore, log(base5)36 ≈ 5.1699/2.3219 ≈ 2.2302.

2. Can I solve log(base5)36 without using a calculator?

Yes, it is possible to solve this problem without a calculator. You can use the logarithm identities, such as log(ab) = log(a) + log(b), to rewrite the problem as log(6^2)/log(5), which equals 2log(6)/log(5). Then, you can use the change of base formula to find log(6)/log(5) ≈ 1.5849, and finally, multiply by 2 to get 2log(6)/log(5) ≈ 3.1699.

3. How can I check my answer for solving log(base5)36 using log2 and log3?

To check your answer, you can use a calculator to evaluate log2(36)/log2(5) and log3(36)/log3(5). Both of these expressions should give you the same result, which is approximately 2.2302. You can also plug this value back into the original equation log(base5)36 and see if it equals to 36.

4. Can I use any other base besides log2 and log3 to solve log(base5)36?

Yes, you can use any other base as long as you are consistent in your calculations. However, using log2 or log3 is most convenient since they are easy to evaluate on a calculator.

5. What is the purpose of using logarithms to solve equations like log(base5)36?

Logarithms are useful because they allow us to solve for unknown variables that are in exponents. In this case, log(base5)36 tells us what exponent we need to raise 5 to in order to get 36. By using logarithms, we can easily find the value of this exponent without having to use trial and error or a lengthy calculation.

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