What is |((e^{x-1}))/(1.5)|<1 ?

The solution to |((e^{x-1}))/(1.5)|<1 is x<ln(1.5)+1

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|((e^{x-1}))/(1.5)|<1

\frac{( e ^{x - 1} )}{1.5} < 1

x < ln (1.5) + 1

Solution steps

\frac{e ^{x - 1}}{1.5} < 1

Apply absolute rule: If

∣ u ∣ < a, a > 0

then

- a < u < a

- 1 < \frac{e ^{x - 1}}{1.5} < 1

\frac{e ^{x - 1}}{1.5} > - 1 and \frac{e ^{x - 1}}{1.5} < 1

True for all x and x < ln (1.5) + 1

Merge Overlapping Intervals

x < ln (1.5) + 1

25 - 9 x^{2} = 0

(x - 3)^{2} = 5

25 x^{2} = 48 - 10 x

x (1 + x) = 6

∣ x - 2 ∣ = 3 x + 1

What is |((e^{x-1}))/(1.5)|<1 ?
The solution to |((e^{x-1}))/(1.5)|<1 is x<ln(1.5)+1