# Exponential Inequalities

**Exponential inequalities** are inequalities in which one (or both) sides involve a variable exponent. They are useful in situations involving repeated multiplication, especially when being compared to a constant value, such as in the case of interest. For instance, exponential inequalities can be used to determine how long it will take to double ones money based on a certain rate of interest; e.g. money will take approximately 12 years to double at a constant interest rate of 6%.

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## Introduction

The key to working with exponential inequalities is the following fact:

If \(a>1\) and \(x>y\), then \(a^x>a^y\). Otherwise, if \(0<a<1\), then \(a^x<a^y\).

Of course, if \(a=1\), then \(1^x=1\) for any \(x\), and similarly if \(a=0\), then \(0^x=0\) for any positive \(x\). More importantly, the converse is true as well:

If \(a>1\) and \(a^x>a^y\), then \(x>y\). Otherwise, if \(0<a<1\) and \(a^x>a^y\), then \(x<y\).

In more formal terms, the exponential function \(f(x)=a^x\) is **monotonically increasing** \(\big(\)increasing \(x\) always increases \(f(x)\big)\) for \(a>1\), and **monotonically decreasing** \(\big(\)increasing \(x\) always decreases \(f(x)\big)\) for \(0<a<1\).

Fortunately, both of these facts are quite intuitive: when the base is greater than 1, the side with the larger exponent will be the greater one, and the opposite is true when the base is less than 1. For instance, without any knowledge of the facts formalized above, one would intuitively expect \(2^{100}\) to be larger than \(2^{95}\).

## Exponential Inequalities - Same Base

When both sides of an inequality have the same base, the key facts from the introduction can be applied directly. For example,

What values of \(x\) satisfy the following inequality: \[2^{2x+3}>2^{3x}?\]

Since the base is 2, which is greater than 1, the fact that \(2^{2x+3}>2^{3x}\) implies that \(2x+3>3x\). Subtracting \(2x\) from both sides gives \(3>x\), so all \(x\) less than \(3\) satisfy the inequality. \(_\square\)

This concept can also be applied to larger "stacks" of exponents:

What values of \(x\) satisfy the following inequality:

\[\large 2^{3^{4x+1}}>2^{3^{2x+3}}?\]

Since the base is 2, which is greater than 1, the given inequality implies \(3^{4x+1} > 3^{2x+3}\). This further implies that \(4x+1>2x+3\), or that \(2x>2 \implies x>1\). Therefore, all \(x\) greater than 1 satisfy the original inequality. \(_\square\)

## Exponential Inequalities - Base less than 1

In the case where the base is less than 1, the previous intuition is essentially reversed: the larger side is now the one with the smaller exponent.

What values of \(x\) satisfy the following inequality: \[\left(\frac{1}{2}\right)^{3x}>\left(\frac{1}{2}\right)^{2x+3}?\]

Since the base is \(\frac{1}{2}\), which is less than 1, the given inequality implies \(3x<2x+3\). Then \(x<3\), so all \(x\) less than 3 satisfy the original inequality. \(_\square\)

## Exponential Inequalities - Similar Base

In many inequalities, the bases are different but can be rewritten in terms of the same base. For example,

What values of \(x\) satisfy the following inequality: \[4^{x+2}>8^{2x}?\]

Here, the bases are different, but they are related by the facts \(4=2^2, 8=2^3\). Rewriting the inequality to use 2 as a base gives

\[2^{2(x+2)}>2^{3(2x)} \implies 2^{2x+4}>2^{6x},\]

so \(2x+4>6x\), implying that \(4>4x\) and \(1>x\). Hence, all \(x\) less than 1 satisfy the original inequality. \(_\square\)

The same concept can be applied when there are several terms:

What values of \(x\) satisfy the following inequality: \[16^x>2^2 \cdot 4^3 \cdot 8^4?\]

Writing all the bases as powers of 2 gives

\[\big(2^4\big)^x>2^2 \cdot \big(2^2\big)^3 \cdot \big(2^3\big)^4 \implies 2^{4x}>2^2 \cdot 2^6 \cdot 2^{12}=2^{20}.\]

Therefore, \(4x>20\implies x>5\). So, all \(x\) greater than 5 satisfy the original inequality. \(_\square\)

## Exponential Inequalities - Different Base

When the two bases are different and not related by a common base (as in the previous section), the use of logarithms becomes necessary. Fortunately, logarithms satisfy essentially the same properties as exponents do:

If \(a>1\) and \(x>y\), then \(\log_ax>\log_ay\). Otherwise, if \(0<a<1\), then \(\log_ax<\log_ay\).

The converse is also true:

If \(a>1\) and \(\log_ax>\log_ay\), then \(x>y\). Otherwise, if \(0<a<1\), then \(x<y\).

The application is generally a matter of straightforward algebra:

What values of \(x\) satisfy the following inequality: \[2^{5x}>5^{8-5x}?\]

Taking the (base 10) logarithm of both sides gives

\[\log 2^{5x} > \log 5^{8-5x} \implies 5x\log 2>(8-5x)\log 5,\]

so \(5x\log 2>8\log 5-5x\log 5.\) Rearranging gives \(5x(\log 2+\log 5)>8\log 5.\) Since \(\log 2+\log 5=\log 10=1,\) this is equivalent to \(5x>8\log 5,\) so \(x>\frac{8}{5}\log 5.\) \(_\square\)

## Exponential Inequalities - Multiple Terms

In the case of multiple terms, it is generally worth assigning another variable to an exponential term, solving the resulting inequality, and then working with the single-term inequality. For example,

What values of \(x\) satisfy the following inequality: \[2^x+4^x>6?\]

Let \(y=2^x\), so that \(y+y^2>6\). This can be rearranged as \(y^2+y-6=(y-2)(y+3)>0\), which is true when \(y>2\) or \(y<-3\). Since \(2^x\) cannot be negative, this implies \(y>2 \implies 2^x>2^1\), which implies \(x>1\). Thus, all \(x\) greater than 1 satisfy the original inequality. \(_\square\)

In the case of an inequality chain, it is usually appropriate to treat each inequality separately, then combine the results. For example,

What values of \(x\) satisfy the following inequality: \[\left(\frac{1}{2}\right)^{x+2}<8<\left(\frac{1}{4}\right)^{2x}?\]

The first inequality is \(\left(\frac{1}{2}\right)^{x+2}<8\), or \(\left(\frac{1}{2}\right)^{x+2}<\left(\frac{1}{2}\right)^{-3}\). Since \(\frac{1}{2}\) is less than 1, this implies that \(x+2>-3\), or that \(x>-5\).

The second inequality is \(8<\left(\frac{1}{4}\right)^{2x}\), or \(\left(\frac{1}{2}\right)^{-3}<\left(\frac{1}{2}\right)^{4x}\). This implies that \(-3>4x\), or that \(-\frac{3}{4}>x\).

Putting the two inequalities together, the solution set is \(-5<x<-\frac{3}{4}\). \(_\square\)

## Exponential Inequalities - Problem Solving

A key strategy is raising both sides of an inequality to the same exponent (usually some fractional exponent, which is the same as taking some root of both sides) in order to simplify the problem:

Find the greatest integer \(x\) for which \[3^{20}>32^x.\]

Since \(32=2^5\), the inequality can be written as \(3^{20}>2^{5x}\implies \big(3^4\big)^5>\big(2^x\big)^5\). This means that \(3^4>2^x\), by taking the \(\frac{1}{5}\) power of both sides of the inequality. Since \(3^4=81\), the largest possible integer is \(6\). Alternatively, \(3^4>2^x\) implies that \(\log_2(3^4)>x \implies 4\log_23>x\), and since \(4\log_23 \approx 6.34\), the largest possible integer value of \(x\) is 6. \(_\square\)

Recall that negative bases only tend to work well with integer exponents; otherwise, the result is usually complex. Therefore, any value that makes a base negative should also make the exponent an integer in order for the expression to remain well-defined.

Solve the inequality \[(x^2+x-2)^{x^2-x-2}>0.\]

We split into cases depending on the value of the base, \( x^2 + x - 2 \).

Case 1:If \(x^2+x-2\) is positive, then the inequality is true whatever the exponent is, so the first case to solve is\[x^2+x-2>0 \implies (x+2)(x-1)>0 \implies x>1 \text{ or } x<-2.\]

Case 2:If \( x^2 + x -2 \) equal 0, then the expression cannot be positive. So there are no solutions.

Case 3:If \(x^2+x-2\) is negative, then \( x \in (-2, 1 ) \).

For \((x^2+x-2)^{x^2-x-2}\) to be positive, \(x^2-x-2\) must be aneveninteger. The values of \( x^2 - x - 2 \) on the restricted domain of \( x \in ( -2, 1 ) \) is \( [ - \frac{9}{4} , 4) \). So the only possible values for \( x^2 - x - 2 \) to be an even integer are when it equals \( -2, 0,\) or \( 2. \).

If \(x^2-x-2=-2\), then \(x^2-x=0 \implies x=0,1\). Since \( 0, 1 \in [ -2, 1 ] \), these are valid solutions.

If \(x^2-x-2=0\), then \((x-2)(x+1)=0 \implies x=-1, 2\). Only \( -1 \in [ -2, 1 ] \), so this is the only solution in this subcase.

If \(x^2-x-2=2\), then \(x^2-x-4=0 \implies x=\frac{1 \pm \sqrt{17}}{2}\). Only \(\frac{1-\sqrt{17}}{2} \in [ -2, 1 ] \), so this is the only solution in this subcase.Combining across the cases, the solution set is

\[x<-2,\ x=\frac{1-\sqrt{17}}{2},\ x=-1,\ x=0,\ x>1.\ _\square\]

## See Also

**Cite as:**Exponential Inequalities.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/exponential-inequalities/