Logarithms

carebear

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Aug 30, 2010
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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?
 
5 = 5, π = π, so aloga(x) = aloga(x)\displaystyle 5 \ = \ 5, \ \pi \ = \ \pi, \ so \ a^{log_a(x)} \ = \ a^{log_a(x)}

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

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

loga(x) = loga(x), loga(a) = 1, hence x = aloga(x)\displaystyle log_a(x) \ = \ log_a(x), \ log_a(a) \ = \ 1, \ hence \ x \ = \ a^{log_a(x)}
 
carebear said:
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:

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

Let

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

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

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

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

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

x = m\displaystyle x \ = \ m

then

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