Solving log(base5)36 using log2 and log3

[seiferseph]

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.

 

[primarygun]

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.

 

[HallsofIvy]

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.

 

[seiferseph]

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.

 

[seiferseph]

sorry, its supposed to be log(base5)36 to solve for

 

[Zurtex]

36 = 3*3*2*2

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

 

[seiferseph]

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?

 

[primarygun]

Yes.
log 5=1-x

 

[Zurtex]

Yes.
log 5=1-x

You sure you have read the edits?

 

[primarygun]

You sure you have read the edits?

What's edited?

 

[Zurtex]

What's edited?

The original question.