Why Monotonicity Matters
When solving an equation such as \(2^x=8\), the value \(x=3\) makes the two sides equal. An inequality asks which values of \(x\) make one side larger or smaller. The key question is whether changing the exponent in an exponential expression changes its value in the same direction.
For a base greater than \(1\), increasing the exponent increases the exponential value. For a base strictly between \(0\) and \(1\), increasing the exponent decreases the value. This difference determines whether an inequality keeps its direction or reverses when we compare exponents.
For an exponential function \(f(x)=a^x\), the base must satisfy \(a>0\). If \(a>1\), the function is increasing; if \(0<a<1\), it is decreasing. The base \(a=1\) gives the constant function \(1^x=1\), which does not have either strict monotonicity.
Why the Base Determines the Direction
Consider two exponents \(x_1<x_2\). Their difference \(h=x_2-x_1\) is positive, and the exponent laws give \(a^{x_2}=a^{x_1}a^h\). If \(a>1\), then \(a^h>1\), so multiplying the positive number \(a^{x_1}\) by \(a^h\) makes it larger. Thus \(a^{x_2}>a^{x_1}\). If \(0<a<1\), then \(a^h<1\), so the product is smaller: \(a^{x_2}<a^{x_1}\).
For example, if \(a>1\), then \(a^u>a^v\) exactly when \(u>v\). If \(0<a<1\), then \(a^u>a^v\) exactly when \(u<v\). The same direction rule applies to \(<,\leq,\) and \(\geq\); equality at a boundary is included only when the original inequality includes equality.
Solving by Matching Bases
If both sides can be written as powers of the same positive base, compare the exponents. First check whether that base is greater than \(1\) or between \(0\) and \(1\). This method is especially direct when one side is a familiar power of the base.
Worked Example: Compare Powers with a Base Greater Than 1
Solve \(2^{x+1}>16\).
Rewrite \(16\) as a power of \(2\): \(16=2^4\). The inequality becomes \(2^{x+1}>2^4\). Since \(2>1\), the exponential function is increasing, so the exponents keep their order:
Subtract \(1\) from both sides to obtain \(x>3\). The solution is \((3,\infty)\). To check the boundary, \(2^{3+1}=2^4=16\), which is not greater than \(16\), so \(x=3\) is correctly excluded. For \(x=4\), the left side is \(2^5=32>16\), confirming that values just above the boundary satisfy the inequality.
Worked Example: Compare Powers with a Base Between 0 and 1
Solve \(\left(\frac{1}{3}\right)^{2x-1}\leq 9\).
Write \(9\) using the base \(\frac{1}{3}\). Since \(\left(\frac{1}{3}\right)^{-2}=3^2=9\), the inequality becomes
The base is positive and less than \(1\), so the exponential function is decreasing. Therefore, the exponent inequality reverses:
Add \(1\) to both sides: \(2x\geq -1\). Divide by the positive number \(2\), which does not change the inequality direction, to get \(x\geq-\frac{1}{2}\). The solution is \(\left[-\frac{1}{2},\infty\right)\). At \(x=-\frac{1}{2}\), the exponent is \(2(-\frac{1}{2})-1=-2\), so both sides equal \(9\); the endpoint belongs in the solution. As an additional check, \(x=0\) gives \(\left(\frac{1}{3}\right)^{-1}=3\leq9\), as expected.
When Matching Bases Is Not Practical
Sometimes the two sides do not have an evident common base. If both sides of an inequality are positive, we can take natural logarithms of both sides because \(\ln\) is increasing. For an inequality \(a^{g(x)}>c\), where \(a>0\) and \(c>0\), taking natural logarithms gives \(g(x)\ln a>\ln c\). The direction has not changed at this step. However, if we then divide by \(\ln a\), the direction changes when \(\ln a<0\), which occurs for \(0<a<1\).
This method requires positivity before taking logarithms. In particular, exponential expressions are positive when their bases are positive, and the constant on the other side must also be positive. If a positive factor multiplies an exponential expression, divide by that factor before taking logarithms; division by a negative factor would itself reverse the inequality.
Worked Example: Use Logarithms with a Base Greater Than 1
Solve \(5^{x-2}<7\).
Both sides are positive, so take natural logarithms. Since \(\ln\) is increasing, the inequality direction stays the same:
Use \(\ln(a^r)=r\ln a\) for \(a>0\):
Because \(5>1\), \(\ln 5>0\). Dividing by \(\ln 5\) keeps the direction:
The solution is \(\left(-\infty,\,2+\frac{\ln 7}{\ln 5}\right)\). At the boundary \(x=2+\frac{\ln 7}{\ln 5}\), the exponent is \(\frac{\ln 7}{\ln 5}\), and \(5^{\ln 7/\ln 5}=e^{\ln 7}=7\). Equality does not satisfy the strict inequality. Since \(5^{x-2}\) is increasing, values below this boundary give values below \(7\), as required.
Worked Example: Use Logarithms with a Base Between 0 and 1
Solve \(\left(\frac{1}{2}\right)^x\geq 6\).
Both sides are positive, so take natural logarithms. The logarithm is increasing and the direction stays the same:
Since \(0<\frac{1}{2}<1\), \(\ln(\frac{1}{2})<0\). Dividing by this negative number reverses the inequality:
The solution is \(\left(-\infty,\,\frac{\ln 6}{\ln(1/2)}\right]\). At the boundary \(x=\frac{\ln 6}{\ln(1/2)}\), we have \(x\ln(\frac{1}{2})=\ln 6\), so \(\left(\frac{1}{2}\right)^x=6\). Equality is included. Because \((\frac{1}{2})^x\) decreases as \(x\) increases, values below the boundary give values greater than \(6\), consistent with the answer.
A Reliable Procedure
Whether you compare bases or take logarithms, make the monotonicity decision explicit. The following sequence prevents most direction errors.
Move terms or divide by a positive coefficient as needed. If dividing by a negative number, reverse the inequality at that step.
For logarithms, both sides must be positive. For comparing exponents, rewrite both sides with one common positive base other than \(1\), if possible.
A base greater than \(1\) preserves exponent order. A base between \(0\) and \(1\) reverses it. With natural logarithms, taking logs preserves order, but dividing by a negative logarithm reverses it.
Keep or exclude the endpoint according to whether the original inequality includes equality. Substitute the boundary into the original expression to verify it.
Common Mistake: Forgetting the Reversal
A frequent error is to compare exponents as if every exponential function were increasing. For example, consider \(\left(\frac{1}{2}\right)^{x-1}>8\). Since \(8=\left(\frac{1}{2}\right)^{-3}\), a mistaken comparison might be written as
This is incorrect because the common base \(\frac{1}{2}\) is between \(0\) and \(1\). The exponential function is decreasing, so the exponent comparison must reverse:
The solution is \((-\infty,-2)\). Checking the boundary gives \(\left(\frac{1}{2}\right)^{-3}=8\), which does not satisfy the strict inequality. Testing \(x=-3\) gives \(\left(\frac{1}{2}\right)^{-4}=16>8\), while \(x=0\) gives \(\frac{1}{2}<8\). These checks agree with \(x<-2\).
Why the Method Works
Matching bases works because a strictly increasing function preserves the order of inputs, while a strictly decreasing function reverses it. Thus, when \(a>1\), applying \(a^x\) to two ordered exponents preserves their order; when \(0<a<1\), it reverses their order. In the logarithm method, taking natural logs preserves order because \(\ln\) is increasing. The sign of \(\ln a\) then determines whether dividing by it preserves or reverses the inequality.
The main practical lesson is to track every operation that can change an inequality direction. The base tells you how exponent comparison behaves; the sign of a divisor tells you whether division reverses the inequality. Checking a boundary and testing a value on the proposed solution side provide useful confirmation.
Check Your Understanding
For each question, identify whether the relevant exponential function is increasing or decreasing, and pay attention to endpoint inclusion.
- Solve \(3^{2x}>27\). Which feature of the base determines whether the exponent comparison reverses?
- Solve \(\left(\frac{1}{4}\right)^{x+1}\leq 16\). Is the boundary included?
- When solving \(2^x>9\) with natural logarithms, why does dividing by \(\ln 2\) keep the inequality direction?
- In the logarithm method for \(\left(\frac{1}{3}\right)^x<5\), which step causes a reversal, if any?
- Why must both sides be positive before taking logarithms of an inequality?
