Logarithmic Inequalities

Logarithmic inequalities are inequalities in which one (or both) sides involve a logarithm. Like exponential inequalities, they are useful in analyzing situations involving repeated multiplication, such as in the cases of interest and exponential decay.

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

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

Of course, the base of a logarithm cannot be 1 or nonpositive. More importantly, the converse is true as well:

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

In more formal terms, the logarithmic function \(f(x)=\log_a x\) 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 argument 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 \(\log_2100\) to be larger than \(\log_295\).

It is also important to keep in mind the following fact:

The argument of a logarithm must be positive!

Thus, it is also necessary to take into account any inequalities resulting from the arguments being positive; for example, an inequality involving the term \(\log_2 (2x-3)\) immediately requires \(x>\frac{3}{2}\).

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:

\[\log_2(2x+3)>\log_2(3x)?\]

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Since the base is 2, which is greater than 1, the fact that \(\log_2(2x+3)>\log_2(3x)\) implies that \(2x+3>3x\). Subtracting \(2x\) from both sides gives \(3>x\).

Additionally, the arguments of both the logarithms must be positive, so additionally \(3x>0\) and \(2x+3>0\). The first is more restrictive, as \(x>0\), so the final solution set is \(0<x<3\). \(_\square\)

This same concept can be applied to larger "stacks" of logarithms:

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

\[\log_2\big(\log_3(4x+1)\big)>\log_2\big(\log_3(2x+3)\big)?\]

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Since the base is 2, which is greater than 1, the given inequality implies \(\log_3(4x+1) > \log_3(2x+3)\). This further implies that \(4x+1>2x+3\), or that \(2x>2 \implies x>1\). Therefore, \(x\) must be greater than 1.

In addition, the arguments of all the logarithms must be positive, but this is indeed the case when \(x\) is greater than 1. Therefore, the solution set is \(x>1\). \(_\square\)

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:

\[\log_{\frac{1}{2}}(3x)>\log_{\frac{1}{2}}(2x+3)?\]

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

Additionally, the argument of each logarithm must be positive, so \(3x>0\) and \(2x+3>0\). The former is more restrictive, so the solution set is \(0<x<3\). \(_\square\)

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:

\[\log_2(x+1)>\log_4\big(x^2\big)?\]

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

\[\log_4\big((x+1)^2\big)>\log_4\big(x^2\big),\]

so \((x+1)^2>x^2\), implying that \(2x+1>0\implies x>-\frac{1}{2}\). Additionally, the arguments of each logarithm must be positive, which excludes the case \(x=0\). Therefore, the final solution set is \(x>-\frac{1}{2}, x \neq 0\). \(_\square\)

This concept can also be applied to constants, by writing \(c\) as \(\log(10^c)\).

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

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

\[\log_2(x)>\log_2(3)+\log_4(25)+\log_8(343)?\]

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Writing all the logarithms in base 2 gives

\[\begin{align} \log_2(x)&>\log_2(3)+\log_2(5)+\log_2(7)\\ \Rightarrow \log_2(x)&>\log_2(105), \end{align}\]

so \(x>105\). All \(x\) greater than 105 satisfy the original inequality. \(_\square\)

When the bases are different and not related by a common base (as in the previous section), the use of the change of base formula becomes necessary. For example,

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

\[ \log_7(x+5)>\log_5(x+5)?\]

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By change of base, the inequality gives

\[\frac{\log(x+5)}{\log 7}>\frac{\log(x+5)}{\log 5}.\]

This is true exactly when \(\log(x+5)\) is negative, meaning that

\[\log(x+5)<0 \implies x+5<1 \implies x<-4,\]

and since \(x+5\) must be positive, \(x>-5\), and the final solution set is \(-5<x<-4\). \(_\square\)

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

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

\[\log_2(x)+\big(\log_2(x)\big)^2>6?\]

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Let \(y=\log_2(x)\), 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 \(y<-3.\)

Thus, either \(\log_2(x)>2 \implies \log_2(x)>\log_24 \implies x>4\), or \(\log_2(x)<-3 \implies \log_2(x)<\log_2\left(\frac{1}{8}\right) \implies x<\frac{1}{8}.\)

The solution set is thus \(0<x<\frac{1}{8}\) and \(x>4\), since the argument of a logarithm must be positive. \(_\square\)

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:

\[\log_{\frac{1}{2}}(x+2)<-2<\log_{\frac{1}{4}}(2x)?\]

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The first inequality is \(\log_{\frac{1}{2}}(x+2)<-2\), or \(\log_{\frac{1}{2}}(x+2)<\log_{\frac{1}{2}}(4)\). Since the base \(\frac{1}{2}\) is less than 1, this implies that \(x+2>4\), so \(x>2\).

The second inequality is \(-2<\log_{\frac{1}{4}}(2x)\), or \(\log_{\frac{1}{4}}(16)<\log_{\frac{1}{4}}(2x)\). Again, the base is less than 1, so this implies \(16>2x\) or \(8>x\).

Thus, \(2<x<8\). Since the argument of each logarithm is positive in this interval, the final solution set is \(2<x<8\). \(_\square\)

Keep in mind that the base of a logarithm can be less than 1, related by the equality

\[\large \log_ab = -\log_{\frac{1}{a}}b,\]

so don't forget this case!

Solve the inequality

\[\log_{\frac{x+4}{2}}\left(\log_{2}\frac{2x-1}{3+x}\right)<0.\]

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Firstly, the base must be positive and not equal to 1, so immediately \(x>-4\) and \(x \neq -2\). Similarly, the argument \(\log_2\frac{2x-1}{3+x}\) must be positive, so \(\frac{2x-1}{x+3}>0 \implies x>\frac12 \text{ or } x<-3\). Thus the only possible values of \(x\) are \(x>\frac12 \, \text{ or }\, -4<x<-3\).

Now consider two cases:

Case 1. \(0<\frac{x+4}{2}<1 \iff -4<x<-2\)

In this case, it is necessary that \(\log_2\frac{2x-1}{3+x}>1\), or \(\frac{2x-1}{3+x}>2\), so \(\frac{2x-1}{3+x}-2=\frac{-7}{3+x}>0\). Hence \(3+x\) is negative, or \(x<-3\). So all \(x\) such that \(-4<x<-3\) satisfy the original inequality.

Remember that this strategy of subtracting the two sides rather than multiplying avoids the casework involved through multiplication, as it is no longer necessary to consider the implications of the multiplied quantity being negative.

Case 2. \(\frac{x+4}{2}>1 \iff x>-2\)

In this case, it is necessary that \(\log_2\frac{2x-1}{3+x}<1\), or \(\frac{2x-1}{3+x}<2\), so \(\frac{2x-1}{3+x}-2=\frac{-7}{3+x}<0\). This is always true as \(x>-2\), so any \(x>-2\) would work. However, recall that only \(x>\frac12 \text{ or } -4<x<-3\) could possibly work, and all \(x\) such that \(x>\frac12\) do.

Therefore, the solution set is

\[x>\frac12,\ -4<x<-3.\ _\square\]

A typical problem-solving strategy is using the change-of-base formula to make all the logarithms have the same base. This makes it much easier to apply other inequalities, such as AM-GM.

Show that

\[\log_n(n+1)>\log_{n+1}(n+2)\]

for all integers \(n \geq 2\).

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Using change of base,

\[\begin{align} \log_n(n+1)>\log_{n+1}(n+2) &\iff \frac{\log (n+1)}{\log n}>\frac{\log (n+2)}{\log (n+1)}\\ &\iff \big(\log (n+1)\big)^2>\log n\log(n+2). \end{align}\]

Products of logarithms are a huge sign to use AM-GM, since the sum of logarithms is very easy to deal with.

In particular,

\[\frac{\log n+\log(n+2)}{2} \geq \sqrt{\log n\log(n+2)} \implies \frac{\log\big(n(n+2)\big)}{2} \geq \sqrt{\log n\log(n+2)}.\]

But \(\log\big(n(n+2)\big)=\log\big(n^2+2n\big)<\log(n^2+2n+1)=\log\big((n+1)^2\big)=2\log(n+1),\) so

\[\log(n+1)>\frac{\log\big(n(n+2)\big)}{2} \geq \sqrt{\log n\log(n+2)}.\]

Therefore,

\[\big(\log(n+1)\big)^2>\log n\log(n+2),\]

proving the original inequality. \(_\square\)

• Logarithms
• Properties of Logarithms - Basic
• Properties of Logarithms - Intermediate
• Exponential Inequalities