Exponent and Logarithm Laws Drive Most Solutions
- Logarithmic equations are usually solved using one (or both) of these strategies:
- Rewrite exponentials and logs into an equivalent form.
- Simplify using exponent and logarithm laws, then solve.
- The most used exponent laws are:
- $a^m\cdot a^n=a^{m+n}$
- $\dfrac{a^m}{a^n}=a^{m-n}$
- $(a^m)^n=a^{mn}$
- $a^{-n}=\dfrac{1}{a^n}$
- $a^{1/n}=\sqrt[n]{a}$
- The matching logarithm laws (for $M>0, N>0$) are:
- $\log_a(MN)=\log_a M+\log_a N$
- $\log_a\left(\dfrac{M}{N}\right)=\log_a M-\log_a N$
- $\log_a(M^k)=k\log_a M$
- The logarithm laws are consequences of exponent laws.
- For example, $MN=a^p\cdot a^q=a^{p+q}$ leads to $\log_a(MN)=p+q=\log_a M+\log_a N$.
Solve Without a Calculator by Recognizing Powers
If you can express the number as an exact power of the base, you can solve instantly.
- $\log_5 625 = x$. Since $625=5^4$, you get $x=4$.
- $\log_4 0.0625 = x$. Since $0.0625=\dfrac{1}{16}=4^{-2}$, you get $x=-2$.
- $\log_2 0.125 = x$. Since $0.125=\dfrac{1}{8}=2^{-3}$, you get $x=-3$.
- $\log_3 1=x$. Since $1=3^0$, you get $x=0$.
Examples with Fractional Exponents
Fractional exponents usually come from roots.
- $\log_5 \sqrt{5}=x$:
- Since $\sqrt{5}=5^{1/2}$, you get $x=\dfrac{1}{2}$.
- $\log_3\left(3\sqrt{3}\right)=x$:
- Rewrite: $3\sqrt{3}=3\cdot 3^{1/2}=3^{3/2}$, so $x=\dfrac{3}{2}$.
- When you see a square root, immediately think "power $1/2$".
- When you see a cube root, think "power $1/3$".
- This often turns a messy log into a one-line solution.
Change of Base Makes Any Logarithm Calculator-Friendly
Change of base formula
For $a>0$, $a\neq 1$, $b>0$, and any convenient base $c>0$, $c\neq 1$:
$$\log_a b=\frac{\log_c b}{\log_c a}$$
Common choices are $c=10$ (using $\log$) or $c=e$ (using $\ln$).
Solve $2^x=50$.
Solution
- Rewrite: $$x=\log_2 50$$
- Use change of base with $c=10$: $$x=\frac{\log 50}{\log 2}\approx 5.64 \text{ (3 s.f.)}$$
- This method is also how you evaluate expressions such as $\log8 5$ or $\log6 121$ to a requested number of decimal places.
Use Logarithms to Solve Exponential Equations
- Many equations have $x$ in the exponent, for example $2^x=50$ or $6^{2x+1}=20$.
- Taking a logarithm converts "exponent problems" into "algebra problems".
General Method (One Unknown Exponent)
To solve $a^{f(x)}=b$:
- Rewrite as $\log_a b=f(x)$, or take logs of both sides.
- Solve the resulting linear equation in $x$.
- If the base is not 10 or $e$, a calculator usually cannot compute $\log_a b$ directly.
- Use the change of base formula to convert it to $\log$ or $\ln$.
Solve $6^{2x+1}=20$.
Solution
- Rewrite in logarithmic form: $$\log_6 20 = 2x+1$$
- Use the change of base formula to evaluate $\log_6 20$ on a calculator, then solve: $$x=\frac{\log_6 20-1}{2} \approx 0.34$$ (rounded to 2 decimal places).
Solve $3^x=40$.
Solution
- Convert to logarithms: $$x=\log_3 40$$
- Then compute with a calculator (using change of base) and round to the required accuracy: $$x \approx 3.36$$
- Do not round too early.
- Keep calculator values in full (or at least 4 to 6 significant figures) until the final step, then round to the accuracy requested.
Log and Exponential Graphs Help You See What Is Happening
Logarithm and exponential functions with the same base are inverse functions, so their graphs are mirror images in the line $y=x$.
Key features to connect to solving equations:
- Solving $2^x=8$ means finding where the exponential graph hits $y=8$.
- Solving $\log_2 x=3$ means finding the $x$-value where the log graph hits $y=3$.
- These solutions match because the functions are inverses.
- Think of $y=2^x$ as a "machine" that takes an input $x$ and outputs a power of 2.
- The logarithm $y=\log_2 x$ is the "undo machine" that takes a power of 2 and returns the exponent you started with.
Solve Exponential Equations by Writing Both Sides with the Same Base
Sometimes you can avoid calculators entirely by expressing both sides as powers of the same base, then equating exponents.
Strategy:
If you can rewrite both sides as $k^{\text{(something)}}$, then: $$k^{A(x)}=k^{B(x)} \Rightarrow A(x)=B(x)$$
Solve $2^{x+1}=4^{2x}$.
Solution
Rewrite $4^{2x}$ using base 2: $$4^{2x}=(2^2)^{2x}=2^{4x}$$
So: $$2^{x+1}=2^{4x} \Rightarrow x+1=4x \Rightarrow x=\frac{1}{3}$$
Solve $\left(\frac{1}{3}\right)^{x+2}=9^{2x-2}$.
Solution
Rewrite both sides in base 3: $$\left(\frac{1}{3}\right)^{x+2}=3^{-(x+2)}$$ $$9^{2x-2}=(3^2)^{2x-2}=3^{4x-4}$$
Equate exponents: $$-(x+2)=4x-4 \Rightarrow -x-2=4x-4 \Rightarrow 2=5x \Rightarrow x=\frac{2}{5}$$
- If you see numbers like $4, 8, 9, 16, 27, 32$, try rewriting them as powers of $2$ or $3$ first.
- This is usually faster than using logs.
Solving Logarithmic Equations Requires Care with Restrictions
A logarithmic equation has the unknown inside a logarithm, for example $\log_2(x-1)=3$.
Basic Pattern: One Logarithm
- If $\log_a(f(x))=c$, convert to exponential form: $$f(x)=a^c$$
- Then solve for $x$, and finally check that $f(x)>0$.
Solve $\log_2(x-1)=3$.
Solution
Convert: $$x-1=2^3=8 \Rightarrow x=9$$
Check: $x-1=8>0$, valid.
When There Are Multiple Logs
You often simplify using log laws.
Solve $\log_a x + \log_a(x-3)=2$.
Solution
- Combine: $$\log_a\big(x(x-3)\big)=2$$
- Convert to exponential: $$x(x-3)=a^2$$
- Then solve the quadratic and check restrictions $x>0$ and $x-3>0$ (so $x>3$).
- Always check restrictions at the end.
- Log equations can produce extraneous solutions when you combine logs or solve after squaring.
- Can you switch correctly between $a^x=b$ and $\log_a b=x$?
- Do you remember the domain rules: base $>0$, base $\neq 1$, argument $>0$?
- Can you solve $k^{A(x)}=k^{B(x)}$ by equating exponents?
- Can you use change of base to evaluate $\log_a b$ on a calculator?
- Do you check solutions against log restrictions before finalizing your answer?
