Log (a ^ k) (m ^ n) = (n / k) log (a) (m) (n ∈ R) how to prove it? Please do not use the bottom changing formula

Log (a ^ k) (m ^ n) = (n / k) log (a) (m) (n ∈ R) how to prove it? Please do not use the bottom changing formula

How to do without changing the bottom formula

What does n in the bottom change formula log [a] B = log [n] B / log [n] a mean and how to calculate it

N represents the new bottom
As for how? I can't answer your question. It's like I can't answer why 1 is equal to 1!
I can only answer you: just change!

Logarithm: what is the difference between log LG ln? Thank you. I checked the book, too Hehe, it's mainly Google's calculator that misled me, LG is set to base 2 and log is set to base 10 I'm not used to it lg(2) = 1 log(10) = 1 ln(2.71828182818281828) = 1

LG is based on 10
Ln is the natural logarithm based on E
Log is added, and the number below is based on that number

Logarithmic correlation: what do log, LG and LN mean respectively?

LG is based on 10
Ln is the natural logarithm based on E
Log is added, and the number below is based on that number

Please tell me some special values of LG, log and Ln ~ (I think so) I forgot everything I learned before~ It's similar to log () = 1. Why can't () in log have special values? Please tell me LG and LN, And trigonometric functions~

The base number of log needs to be given
LG base 10
The base number of LN is e
() cannot be non positive
lg10=1
lg1=0
lne=1
ln1=0
-----------------

How to write log, LN and LG in C language Suppose the base number in the log is 5. The other assumption is 100. Write it as log5 (100) But how to write ln100 and LG100? PS: ln is based on e? Is the base number of LG 10?

C directly provides the natural logarithm log with E as the base and the common logarithm log10 with 10 as the base
Other logarithms can be written as a function
#include
#include
double loga(double n,double base);
int main (void)
{
double a,b,c;
a = log(exp(1));
b = log10(10);
c = loga(100,5);
printf("%lf %lf %lf",a,b,c);
}
double loga(double n,double base)
{
return log(n) / log(base);
}