How Do You Solve Exponential Inequalities Using Logarithms?

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SUMMARY

The discussion focuses on solving the exponential inequality \(15(5^x - 3^x) < 16 \cdot 15^{\frac{x}{2}}\). Participants suggest rearranging the equation to \(5^x - 3^x < 16 \cdot 15^{\frac{x-2}{2}}\) and recommend dividing by \(3^x\) to simplify the terms. This leads to a comparison of \(\left(\frac{5}{3}\right)^x\) and \(\left(\frac{5}{3}\right)^{x/2}\), establishing that the first term is the square of the second. The use of logarithmic rules is essential for further manipulation of the inequality.

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Homework Statement


##15(5^x-3^x)<16⋅15^{\frac{x}{2}}##

Homework Equations


Rules of logarithms

The Attempt at a Solution


I don't know where to start.

Here's one way to start rearranging the equation:
##5^x-3^x<16⋅15^{\frac{x-2}{2}}##

What's the correct way to start rearranging?
 
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kaspis245 said:

Homework Statement


##15(5^x-3^x)<16⋅15^{\frac{x}{2}}##

Homework Equations


Rules of logarithms

The Attempt at a Solution


I don't know where to start.

Here's one way to start rearranging the equation:
##5^x-3^x<16⋅15^{\frac{x-2}{2}}##

What's the correct way to start rearranging?
I suggest dividing by 3x .

Added in Edit:
Then you have one term with ##\displaystyle \ \left(\frac{5}{3}\right)^x \ ## and another with ##\displaystyle \ \left(\frac{5}{3}\right)^{x/2} \ ## , the first being the square of the second.
 
Last edited:

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