How Do You Solve Exponential Inequalities Using Logarithms?

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To solve the exponential inequality 15(5^x - 3^x) < 16·15^(x/2), start by rearranging the equation to isolate terms involving x. One approach is to express it as 5^x - 3^x < 16·15^((x-2)/2). Dividing by 3^x can simplify the terms, leading to a comparison between (5/3)^x and (5/3)^(x/2). This transformation highlights the relationship between the terms, making it easier to analyze the inequality. Understanding these steps is crucial for effectively solving exponential inequalities using logarithms.
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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:
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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