Are there extraneous roots in this?

[metsandnets]

I need to find the solution set of e^(2x - 3) < e^-1. Can I just reduce this down to 2x - 3 < -1? Thanks.

 

[Deleted member 4993]

I need to find the solution set of e^(2x - 3) < e^-1. Can I just reduce this down to 2x - 3 < -1? Thanks.

Yes... only one solution.

You confirm it by plotting

y = e^2x-3

and

y = e^-1

and checking for point/s of intersection.

 

[pka]

I need to find the solution set of e^(2x - 3) < e^-1. Can I just reduce this down to 2x - 3 < -1? Thanks.

What is your question?

 

[metsandnets]

What is the solution set of e^(2x - 3) < e^(-1) ?

 

[mmm4444bot]

I'm thinking that the meaning of Subhotosh's "only one solution" is that the solution set for the original exercise is the same as the solution set for the "reduced" inequality posted.

In other words, you may go ahead and solve the simpler one.

NOTE: Working with the original inequality, instead, is not that involved, for those familiar with properties of exponents and natural logarithms.

 

[HallsofIvy]

What is the solution set of e^(2x - 3) < e^-1?

$\displaystyle e^{2x-3}= \frac{e^{2x}}{e^3}= \frac{(e^x)^2}{e^3}$ so that $\displaystyle e^{2x-3}= e^{-1}$ is the same as $\displaystyle \frac{(e^x)^2}{e^3}= e^{-1}$ which is the same as $\displaystyle (e^x)^2= \frac{e^3}{e}= e^2$. Taking the square root of each side, $\displaystyle e^x= e$ so that x= 1. Now, because all functions here are continuous, x= 1 is the only place where "<" can change to ">". Check one value of x less than 1 and one value of x greater than 1 to see which satisfies the inequality.

As for your original question, yes, the natural logarithm function is an increasing function so if a< b then ln(a)< ln(b). It is perfectly correct to take the logarithm of both sides and say that 2x- 3< -1 so that 2x< 2, x< 1.