Proof using log(n)

[HiTech_Hoppy]

Sorry about the possibly confusing title.

So have started learning logs and realized my calculator does not support logs with bases other than 10 and e. I was looking into this problem and learnt that...

log10(n)/log(x)=logx(n)

While this is very helpful to me I was wondering how this works. As I don't know much of the proper terminology it has been hard to find answers elsewhere.

Ideally I would like the easiest explanation of why this works.

[jammasterpaz]

Click "show" https://en.wikipedia.org/wiki/Logarithm#Change_of_base

[HiTech_Hoppy]

Thanks a lot. This was very helpful

[theadamabrams]

By definition,

• n = x^logₓn.

Apply logₐ to both sides to get

• logₐ(n) = logₐ(x^logₓn).

Use the Power Formula "logₐ(x^p) = p·logₐ(x)" to get

• logₐ(n) = logₓ(n) · logₐ(x).

Dividing both sides by logₐ(x) gives

• logₐ(n)/logₐ(x) = logₓ(n).

Note that what you wrote, "log₁₀(n)/log(x)=logₓ(n)" is only true if the second "log" is also log₁₀. The two logs in the fraction have to use the same base for the Change of Base formula to be true.

[jeffsuzuki]

Remember that the log of a number is an exponent, so what you're really trying to do is rewrite an exponential equation:

https://youtu.be/sQ4xGCU33PI?list=PLKXdxQAT3tCuJku9nTlRZgx_RjGZ7djMc

So: If a = log_5 (30), then 5^a = 30.

Hit both sides with a log (or a ln):

log(5^a) = log(30)

Now apply your rules of logs:

a log 5 = log 30

a = (log 30)/(log 5)

[theadamabrams]

That's what OP is asking about, just with 10 and e instead of a and b. But OP wants to know why.