Solve the inequality graphically: 7^x > 3^x?
回答 (6)
2016-03-15 2:46 pm
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
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
2016-03-15 2:34 pm
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
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
2017-03-27 9:53 pm
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
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
2016-11-29 6:49 pm
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
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
2016-08-10 9:16 pm
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
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
2016-03-15 3:41 pm
x>0
2016-03-15 2:49 pm
2016-03-15 2:42 pm
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.
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.
2016-03-15 2:38 pm
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
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
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