Practice Exponential Inequality 3

MarkFL

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Write the RHS as a power of 5...
 

MarkFL

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Yes.
 

MarkFL

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You can't just replace 125 with 5...what you must do is as follows:

\(\displaystyle 5^{2x}<125^{x-5}\)

\(\displaystyle 5^{2x}<(5^3)^{x-5}\)

\(\displaystyle 5^{2x}<5^{3(x-5)}\)

\(\displaystyle 5^{2x}<5^{3x-15}\)
 
Jan 15, 2020
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You can't just replace 125 with 5...what you must do is as follows:

\(\displaystyle 5^{2x}<125^{x-5}\)

\(\displaystyle 5^{2x}<(5^3)^{x-5}\)

\(\displaystyle 5^{2x}<5^{3(x-5)}\)

\(\displaystyle 5^{2x}<5^{3x-15}\)
I made a typo. I know that 125 = 5^3.
 
Jan 15, 2020
659
7
18
54
Bronx, NY
You can't just replace 125 with 5...what you must do is as follows:

\(\displaystyle 5^{2x}<125^{x-5}\)

\(\displaystyle 5^{2x}<(5^3)^{x-5}\)

\(\displaystyle 5^{2x}<5^{3(x-5)}\)

\(\displaystyle 5^{2x}<5^{3x-15}\)
2x < 3x - 15

2x - 3x < - 15

-x < - 15

x < - 15/-1

x < 15
 

MarkFL

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When you multiply/divide both sides of an inequality by a negative number, the direction of the inequality must change.
 
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Jan 15, 2020
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Bronx, NY
When you multiply/divide both sides of an inequality by a negative number, the direction of the inequality must change.
Yes. I forgot this rule.

2x < 3x - 15

2x - 3x < - 15

-x < - 15

x > - 15/-1

x > 15
 
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MarkFL

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Yes, that's correct. :)