Logarithms

[carebear]

I understand that log base a of a ^ x is equal to x because of the power law of logarithms......x multiplied by log base a of a which is x multiplied by 1, which is x....I understand that one.....BUT....

I know that a^(log base a of x) = x ....I just don't understand why....can you please explain it mathematically?

 

[BigGlenntheHeavy]

\(\displaystyle 5 \ = \ 5, \ \pi \ = \ \pi, \ so \ a^{log_a(x)} \ = \ a^{log_a(x)}\)

\(\displaystyle Hence, \ by \ the \ law \ of \ logs, \ log_a{[a^{log_a(x)}] \ = \ log_a{[a^{log_a(x)}].\)

\(\displaystyle [log_a(x)][log_a(a)] \ = \ [log_a(x)][log_a(a)]\)

\(\displaystyle log_a(x) \ = \ log_a(x), \ log_a(a) \ = \ 1, \ hence \ x \ = \ a^{log_a(x)}\)

 

[Deleted member 4993]

I understand that log base a of a ^ x is equal to x because of the power law of logarithms......x multiplied by log base a of a which is x multiplied by 1, which is x....I understand that one.....BUT....

I know that a^(log base a of x) = x ....I just don't understand why....can you please explain it mathematically?

It is correct that:

\(\displaystyle Log_a(a^x) \ = \ x\) ........................................................(1)

Let

\(\displaystyle a^{Log_a(x)} \ = \ m\)

then if we take Log[sub:npxyqbsn]a[/sub:npxyqbsn] on both sides we get:

\(\displaystyle Log_a(a^{[Log_a(x)]}) \ = \ Log_a(m)\)

\(\displaystyle [Log_a(x)] * Log_a(a)\ = \ Log_a(m)\)

\(\displaystyle [Log_a(x)] * 1\ = \ Log_a(m)\)

\(\displaystyle x \ = \ m\)

then

\(\displaystyle a^{Log_a(x)} \ = \ m \ = \ x\)