SOLUTION: Compare equations y=log5x(the 5 is at the bottom next to the g) & y=5^x; y=log1/3x(the 1/3 is low next to the g) & y=(1/3)^x; y=log5x & y=log(1/3)x (both the 5 and 1/3 are low next

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Compare equations y=log5x(the 5 is at the bottom next to the g) & y=5^x; y=log1/3x(the 1/3 is low next to the g) & y=(1/3)^x; y=log5x & y=log(1/3)x (both the 5 and 1/3 are low next      Log On


   

Question 349105: Compare equations y=log5x(the 5 is at the bottom next to the g) & y=5^x; y=log1/3x(the 1/3 is low next to the g) & y=(1/3)^x; y=log5x & y=log(1/3)x (both the 5 and 1/3 are low next to the g)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+log%285%2C+%28x%29%29
(Note: The right side is called "the base 5 log of x".)
y+=+5%5Ex
If we rewrite the first equation in exponential form we get:
5%5Ey+=+x
These two equation are identical except the x's and y's have traded places. This makes the two equations inverses of each other.

y+=+log%281%2F3%2C+%28x%29%29
y+=+%281%2F3%29%5Ex
Again, if we rewrite the first equation in exponential form we get:
%281%2F3%29%5Ey+=+x
Again we have inverses.

y+=+log%285%2C+%28x%29%29
y+=+log%281%2F3%2C+%28x%29%29
Other than these both being logarithmic functions, there is no special relationship between these two equations.