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

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

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

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

I made a typo. I know that 125 = 5^3.

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

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

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

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

2x < 3x - 15

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

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

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

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

2x - 3x < - 15

-x < - 15

x < - 15/-1

x < 15

Yes. I forgot this rule.

2x < 3x - 15

2x - 3x < - 15

-x < - 15

x > - 15/-1

x > 15

This inequality math here reminds me of algebra 1?Yes, that's correct.