Loagrithms and -log

[OhMrsDarcy]

The problem asks:

Show that if a<1 that loga(X)= -log1/a(X).

This is random problem our teacher assigned us not out of the book, and there is nothing in our book about negative logs. Anyone have any clues?

 

[pka]

\(\displaystyle \L
\begin{array}{l}
y = \log _a (x) \\
a^y = x \\
a^{ - y} = \frac{1}{x} \\
\left( {a^{ - 1} } \right)^y = \frac{1}{x} \\
\log _{a^{ - 1} } \left( {\frac{1}{x}} \right) = y \\
- \log _{\frac{1}{a}} (x) = y \\
\end{array}\)

 

[Mrspi]

The problem asks:
Show that if a<1 that loga(X)= -log1/a(X).
This is random problem our teacher assigned us not out of the book, and there is nothing in our book about negative logs. Anyone have any clues?

Ok....let

log<SUB>a</SUB> x = n

Convert to exponential form:

a<SUP>n</SUP> = x

Take the reciprocal of both sides:

1 / a<SUP>n</SUP> = 1 / x

Now, 1 / a<SUP>n</SUP> is the same thing as (1 / a)<SUP>n</SUP>, and 1 / x is the same thing as x<SUP>-1</SUP>. So, we can write

(1 / a)<SUP>n</SUP> = x<SUP>-1</SUP>

Change this back to logarithmic notation....note that the base is now (1/a):

log<SUB>(1/a)</SUB> x<SUP>-1</SUP> = n

Use the rule for logs which says that log<SUB>b</SUB> a<SUP>m</SUP> = m log<SUB>b</SUB> a:

-1 * log<SUB>(1/a)</SUB> x = n

Now, since both log<SUB>a</SUB> x and -1 * log<SUB>(1/a)</SUB> x are equal to n, they must be equal to each other:

log<SUB>a</SUB> x = - log<SUB>(1/a)</SUB> x

 

[OhMrsDarcy]

OH! Thank you both so much! That helps with all the other problems as well!