Solve the inequality graphically: 7^x > 3^x?

You are asked to solve it graphically, so you would need to graph each function (see the link below):
f(x) = 7^x
g(x) = 3^x

When x is positive (x > 0), the graph of the first function (7^x) is above the second function (3^x)
When x is zero (x = 0), they are both 1.
When x is negative (x < 0), the graph of the first function (7^x) is below the second function (3^x)

So the only place where this inequality would be true is when x > 0. You can also solve this algebraically, but I'm sure you've already been shown how to do that.

Answer:
x > 0

7^x = 3^x

log(7^x) = log(3^x)

x log(7) = x log(3)

x log(7) - x log(3) = 0

x [log(7) - log(3)] = 0

x = 0

7^(x) &amp;amp;gt; 3^(x)

ln[7^(x)] &amp;amp;gt; ln[3^(x)] → recall: ln(x^a) = a...........................ln(x)

x...........................ln(7) &amp;amp;gt; x...........................ln(3)

x...........................ln(7) - x...........................ln(3) &amp;amp;gt; 0

x...........................[ln(7) - ln(3)] &amp;amp;gt; 0 → as you know that: [ln(7) - ln(3)] &amp;amp;gt; 0

x &amp;amp;gt; 0

7^(x) &amp;gt; 3^(x)

ln[7^(x)] &amp;gt; ln[3^(x)] → recall: ln(x^a) = a.........ln(x)

x.........ln(7) &amp;gt; x.........ln(3)

x.........ln(7) - x.........ln(3) &amp;gt; 0

x.........[ln(7) - ln(3)] &amp;gt; 0 → as you know that: [ln(7) - ln(3)] &amp;gt; 0

x &amp;gt; 0

7^(x) &gt; 3^(x)

ln[7^(x)] &gt; ln[3^(x)] → recall: ln(x^a) = a...ln(x)

x...ln(7) &gt; x...ln(3)

x...ln(7) - x...ln(3) &gt; 0

x...[ln(7) - ln(3)] &gt; 0 → as you know that: [ln(7) - ln(3)] &gt; 0

x &gt; 0

x>0

x > 0
http://www.wolframalpha.com/input/?i=graph+7%5Ex+and+3%5Ex+from+-2+to+1

7^x > 3^x

x ln 7 > x ln 3

Now if x > 0, we get

ln 7 > ln 3, which is true

for any x > 0.

If x = 0, we get 0 > 0, which

is impossible.

If x < 0, we get ln 7 < ln 3, which is false.

So the solution is all x > 0. Graphically, it's

the number line to the right of 0.

7^(x) > 3^(x)

Ln[7^(x)] > Ln[3^(x)] → recall: Ln(x^a) = a.Ln(x)

x.Ln(7) > x.Ln(3)

x.Ln(7) - x.Ln(3) > 0

x.[Ln(7) - Ln(3)] > 0 → as you know that: [Ln(7) - Ln(3)] > 0

x > 0