Logarithmic equations and inequalities

Exponential and logarithmic functions, grade 12 (5 units) Lesson 5 of 23Learning

Logarithmic equations and inequalities

The domain first

A logarithm is defined only for a positive number. So a logarithmic equation is solved in its domain, and you find the domain before you start:

In $\displaystyle \log_{2}(x+2)+\log_{2}(x-2)$ you need $\displaystyle x+2>0$ and $\displaystyle x-2>0$ , that is $\displaystyle x>2$ .
The laws of logarithms can widen the domain: $\displaystyle \log_{2}\left((x+2)(x-2)\right)$ is defined for $\displaystyle x<-2$ too. A solution outside the domain of the original equation is rejected.

One logarithm on each side

$\displaystyle \log_a x$ increases or decreases, so it takes every value once. When each side holds one logarithm to the same base, or a logarithm and a number:

\displaystyle \log_{a}f(x)=\log_{a}g(x)\iff f(x)=g(x)

\displaystyle f(x)>0,\ g(x)>0

\displaystyle \log_{a}f(x)=c\iff f(x)=a^{c}

Worked example: a sum of logarithms

The example below solves $\displaystyle {\log_{2}(x+2)+\log_{2}(x-2)}={5}$ :

The domain: $\displaystyle x>2$ .
The logarithm of a product: $\displaystyle {\log_{2}(x^{2}-4)}={5}$ , so $\displaystyle {x^{2}-4}={2^{5}}={32}$ .
$\displaystyle {x^{2}}={36}$ , that is $\displaystyle {x}={6}$ or $\displaystyle {x}={-6}$ .
$\displaystyle -6$ is outside the domain. The solution: $\displaystyle {x}={6}$ .

Different bases

When the bases differ, change all the logarithms to one base. The example below solves $\displaystyle {\log_{3}(x-1)+\log_{9}(x-1)}={3}$ , in the domain $\displaystyle x>1$ :

$\displaystyle {\log_{9}(x-1)}={\frac{\log_{3}(x-1)}{\log_{3}9}}={\frac{1}{2}\log_{3}(x-1)}$ .
$\displaystyle {\frac{3}{2}\log_{3}(x-1)}={3}$ , so $\displaystyle {\log_{3}(x-1)}={2}$ .
$\displaystyle {x-1}={9}$ , and $\displaystyle {x}={10}$ .

The substitution

\displaystyle {t}={\log_a x}

When an equation holds $\displaystyle \log_a x$ and $\displaystyle (\log_a x)^2$ , substitute $\displaystyle {t}={\log_a x}$ . The example below solves $\displaystyle {(\log_{2}x)^{2}-\log_{2}x-6}={0}$ , in the domain $\displaystyle x>0$ :

$\displaystyle {t^{2}-t-6}={0}$ , so $\displaystyle {t}={3}$ or $\displaystyle {t}={-2}$ .
$\displaystyle {\log_{2}x}={3}$ gives $\displaystyle {x}={8}$ ; $\displaystyle {\log_{2}x}={-2}$ gives $\displaystyle {x}={\frac{1}{4}}$ .
$\displaystyle t$ can be negative: $\displaystyle \log_a x$ takes every value.

An exponential equation with different bases

When the two sides cannot be written as powers of the same base, take the logarithm of both sides. The example below solves $\displaystyle {2^{x}}={5^{x-1}}$ :

$\displaystyle {\log2^{x}}={\log5^{x-1}}$ , that is $\displaystyle {x\log2}={(x-1)\log5}$ .
$\displaystyle {x\log5-x\log2}={\log5}$ , so $\displaystyle {x}={\frac{\log5}{\log5-\log2}\approx1.756}$ .

Logarithmic inequalities

As in an exponential inequality: for $\displaystyle a>1$ the sign stays, and for $\displaystyle 0<a<1$ it turns – and in addition, the solution lies inside the domain. The example below solves $\displaystyle \log_{\frac{1}{2}}(x-1)>-2$ :

The domain: $\displaystyle x>1$ .
$\displaystyle {-2}={\log_{\frac{1}{2}}4}$ , because $\displaystyle {\left(\frac{1}{2}\right)^{-2}}={4}$ .
The base $\displaystyle \frac{1}{2}$ is less than 1, so the sign turns: $\displaystyle x-1<4$ , that is $\displaystyle x<5$ .
Together with the domain: $\displaystyle 1<x<5$ .

Exercise 1

Solve.

Exercise 2

Solve with the change of base.

Exercise 3

Solve with a substitution.

Exercise 4

Solve by taking the logarithm to base 10 of both sides.

Exercise 5

Solve the inequalities.

Formulas: exponential and logarithmic functions, grade 12 (5 units)

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Powers

A product of powers with the same base
$\displaystyle a^{x}\cdot a^{y}$ $\displaystyle =$ $\displaystyle a^{x+y}$
A quotient of powers with the same base
$\displaystyle \frac{a^{x}}{a^{y}}$ $\displaystyle =$ $\displaystyle a^{x-y}$
A power of a power
$\displaystyle (a^{x})^{y}$ $\displaystyle =$ $\displaystyle a^{x\cdot y}$
A power of a product
$\displaystyle (a\cdot b)^{x}$ $\displaystyle =$ $\displaystyle a^{x}\cdot b^{x}$
A power of a quotient
$\displaystyle \left(\frac{a}{b}\right)^{x}$ $\displaystyle =$ $\displaystyle \frac{a^{x}}{b^{x}}$
A negative exponent
$\displaystyle a^{-x}$ $\displaystyle =$ $\displaystyle \frac{1}{a^{x}}$
The exponent 0
$\displaystyle a^{0}$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle a\neq0$
A rational exponent
$\displaystyle a^{\frac{m}{n}}$ $\displaystyle =$ $\displaystyle \sqrt[n]{a^{m}}$ $\displaystyle a>0$

Logarithms (within the domain)

The definition of the logarithm
$\displaystyle \log_{a}x=b\iff a^{b}=x$ $\displaystyle a>0,\ a\neq1,\ x>0$
$\displaystyle \log_{a}(a^{b})$ $\displaystyle =$ $\displaystyle b$
$\displaystyle a^{\log_{a}x}$ $\displaystyle =$ $\displaystyle x$
The logarithm of a product
$\displaystyle \log_{a}x+\log_{a}y$ $\displaystyle =$ $\displaystyle \log_{a}(x\cdot y)$
The logarithm of a quotient
$\displaystyle \log_{a}x-\log_{a}y$ $\displaystyle =$ $\displaystyle \log_{a}\left(\frac{x}{y}\right)$
The logarithm of a power
$\displaystyle \log_{a}(x^{b})$ $\displaystyle =$ $\displaystyle b\cdot\log_{a}x$
Change of base
$\displaystyle \log_{c}x$ $\displaystyle =$ $\displaystyle \frac{\log_{a}x}{\log_{a}c}$
$\displaystyle \log_{a}b$ $\displaystyle =$ $\displaystyle \frac{1}{\log_{b}a}$
The natural logarithm – to the base $\displaystyle e$
$\displaystyle \ln x=\log_{e}x$

The exponential and the logarithmic function

Every exponential as a power of $\displaystyle e$
$\displaystyle a^{x}$ $\displaystyle =$ $\displaystyle e^{x\ln a}$
$\displaystyle a^x$ and $\displaystyle \log_a x$ – inverse functions
$\displaystyle y=a^{x}\iff x=\log_{a}y$ $\displaystyle y>0$
The number $\displaystyle e$
$\displaystyle e\approx2.718$

Growth and decay

The amount after $\displaystyle t$ units of time
$\displaystyle f(t)=f(0)\cdot q^{t}$
The growth / decay factor – $\displaystyle p$ the percent per unit of time
$\displaystyle q=1\pm\frac{p}{100}$
The half-life $\displaystyle T$
$\displaystyle q^{T}$ $\displaystyle =$ $\displaystyle \frac{1}{2}$
The doubling time $\displaystyle T$
$\displaystyle q^{T}$ $\displaystyle =$ $\displaystyle 2$

Derivatives

$\displaystyle (e^{x})^{\prime}$ $\displaystyle =$ $\displaystyle e^{x}$
$\displaystyle (a^{x})^{\prime}$ $\displaystyle =$ $\displaystyle a^{x}\cdot\ln a$
$\displaystyle (\ln x)^{\prime}$ $\displaystyle =$ $\displaystyle \frac{1}{x}$
$\displaystyle (\log_{a}x)^{\prime}$ $\displaystyle =$ $\displaystyle \frac{1}{x\cdot\ln a}$
The derivative of $\displaystyle e^{f(x)}$
$\displaystyle \left[e^{f(x)}\right]^{\prime}$ $\displaystyle =$ $\displaystyle e^{f(x)}\cdot f^{\prime}(x)$
The derivative of $\displaystyle \ln(f(x))$
$\displaystyle \left[\ln(f(x))\right]^{\prime}$ $\displaystyle =$ $\displaystyle \frac{f^{\prime}(x)}{f(x)}$
The derivative of a product
$\displaystyle \left[f(x)\cdot g(x)\right]^{\prime}$ $\displaystyle =$ $\displaystyle f^{\prime}(x)\cdot g(x)+f(x)\cdot g^{\prime}(x)$
The derivative of a quotient
$\displaystyle \left[\frac{f(x)}{g(x)}\right]^{\prime}$ $\displaystyle =$ $\displaystyle \frac{f^{\prime}(x)\cdot g(x)-f(x)\cdot g^{\prime}(x)}{\left[g(x)\right]^{2}}$
The derivative of a composite function
$\displaystyle \left[f(u(x))\right]^{\prime}$ $\displaystyle =$ $\displaystyle f^{\prime}(u)\cdot u^{\prime}(x)$
The tangent’s equation at the point where $\displaystyle {x}={a}$
$\displaystyle y-f(a)=f^{\prime}(a)(x-a)$

Integrals, the area and accumulation

$\displaystyle \int e^{x}\,dx=e^{x}+C$
$\displaystyle \int a^{x}\,dx=\frac{a^{x}}{\ln a}+C$
In any interval where $\displaystyle x\neq0$
$\displaystyle \int\frac{1}{x}\,dx=\ln|x|+C$
$\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int f(mx+b)\,dx=\frac{1}{m}F(mx+b)+C$ $\displaystyle m\neq0$
$\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int f(u(x))\cdot u^{\prime}(x)\,dx=F(u(x))+C$
Where $\displaystyle f(x)\neq0$
$\displaystyle \int\frac{f^{\prime}(x)}{f(x)}\,dx=\ln|f(x)|+C$
$\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int_{a}^{b}f(x)\,dx=F(b)-F(a)$
The area between two graphs
$\displaystyle S=\int_{a}^{b}\left(f(x)-g(x)\right)dx$ $\displaystyle f(x)\geq g(x)$
The accumulation function’s derivative
$\displaystyle \left(\int_{a}^{x}f(t)\,dt\right)^{\prime}=f(x)$