Question 1048381
<pre><b>
{{{drawing(400,1580/3,-3,3,-1.9,6,grid(1),
graph(400,1580/3,-3,3,-1.9,6,3^x),red(line(-5,5,5,5)) )}}} 

We draw a green vertical line from the point where y=5 on the
red line where the graph of y = 3<sup>x</sup> crosses it:

{{{drawing(400,1580/3,-3,3,-1.9,6,grid(1),green(line(1.464973521,0,1.464973521,5)),
graph(400,1580/3,-3,3,-1.9,6,3^x),red(line(-5,5,5,5)) )}}}

Then we look down at the x-axis and see that the green line
touches the x-axis at a little bit less than 1.5, maybe about 
1.46.  It's actually 1.464973521 which can be calculated with
logarithms.

{{{3^x}}}{{{""=""}}}{{{5}}}

Take the logs of both sides:

{{{log((3^x))}}}{{{""=""}}}{{{log((5))}}}

Use the property of logs {{{log((B^A))=B*log((A))}}} 

{{{x*log((3))}}}{{{""=""}}}{{{log((5))}}}

Divide both sides by log(3)

{{{x}}}{{{""=""}}}{{{log((5))/log((3))}}}{{{""=""}}}{{{0.6989700043/0.4771212547}}}{{{""=""}}}{{{1.464973521}}} 

If we do the same with the graph of y = 3<sup>-x</sup>, it just
puts it "in the mirror", i.e., reflects across the y-axis, and
the value of x is just the negative of what it was above.

{{{drawing(400,1580/3,-3,3,-1.9,6,grid(1),green(line(-1.464973521,0,-1.464973521,5)),
graph(400,1580/3,-3,3,-1.9,6,1/3^x),red(line(-5,5,5,5)) )}}} 

Edwin</pre></b>