Exponential and logarithmic equations are equations where the variable appears in an exponent,
inside a logarithm, or both.
An exponential equation may look like:
\[
2^{x+1}=3^x
\]
A logarithmic equation may look like:
\[
\log_3(x-2)+4=6
\]
1. Solving exponential equations
If both sides can be written with the same base, use the one-to-one property of exponential functions.
\[
b^u=b^v
\]
Then:
\[
u=v
\]
For example:
\[
2^{x+1}=2^5
\]
Since the bases are the same, the exponents must be equal:
\[
x+1=5
\]
Therefore:
\[
x=4
\]
2. When bases are different
If the bases are different, logarithms are often needed. For example:
\[
2^{x+1}=3^x
\]
Take the natural logarithm of both sides:
\[
\ln\left(2^{x+1}\right)=\ln\left(3^x\right)
\]
Use the power rule:
\[
(x+1)\ln(2)=x\ln(3)
\]
Expand and collect the \(x\)-terms:
\[
x\ln(2)+\ln(2)=x\ln(3)
\]
\[
\ln(2)=x\ln(3)-x\ln(2)
\]
\[
\ln(2)=x\left(\ln(3)-\ln(2)\right)
\]
Solve for \(x\):
\[
x=\frac{\ln(2)}{\ln(3)-\ln(2)}
\]
Numerically:
\[
x\approx1.710
\]
3. Solving logarithmic equations
A logarithmic equation can often be solved by converting the logarithmic form into exponential form.
\[
\log_b(M)=N
\]
This means:
\[
b^N=M
\]
For example:
\[
\log_3(x-2)+4=6
\]
First isolate the logarithm:
\[
\log_3(x-2)=2
\]
Convert to exponential form:
\[
3^2=x-2
\]
Therefore:
\[
9=x-2
\]
\[
x=11
\]
4. Domain checking
Logarithms require positive arguments. This is the most important restriction in logarithmic equations.
\[
\log_b(M)
\]
is real only when:
\[
M>0
\]
For the equation:
\[
\log_3(x-2)+4=6
\]
the argument is \(x-2\), so the domain condition is:
\[
x-2>0
\]
Therefore:
\[
x>2
\]
The solution \(x=11\) is valid because \(11>2\).
5. Extraneous solutions
An extraneous solution is a value that appears during algebraic solving but does not satisfy the
original equation.
This can happen when logarithms are combined, when both sides are exponentiated, or when both sides are
squared.
Every candidate solution must be substituted back into the original equation.
A valid solution must pass two checks:
- All logarithm arguments must be positive.
- The left side and right side must be equal.
6. Graph intersection method
Any equation can be written as:
\[
L(x)=R(x)
\]
The graph intersection method plots the left side \(L(x)\) and the right side \(R(x)\).
Every intersection point gives a solution.
Another equivalent method is to define:
\[
F(x)=L(x)-R(x)
\]
Then solutions occur when:
\[
F(x)=0
\]
On the graph, this means the difference curve crosses the \(x\)-axis.
7. Numerical solving
Some equations cannot be solved neatly by hand. For example:
\[
e^x=x+3
\]
This equation mixes an exponential expression with a linear expression. In such cases, a numerical method
is useful.
The calculator scans the selected interval for sign changes in:
\[
F(x)=L(x)-R(x)
\]
If \(F(x)\) changes sign between two nearby \(x\)-values, a root lies between them. The calculator then
refines the root using bisection.
8. Bisection idea
Suppose:
\[
F(a)<0
\]
and:
\[
F(b)>0
\]
Then a root exists somewhere between \(a\) and \(b\), provided \(F\) is continuous on that interval.
The midpoint is:
\[
m=\frac{a+b}{2}
\]
The method keeps the half-interval where the sign change remains, and repeats until the root is accurate.
9. Common solving rules
| Situation |
Rule |
Meaning |
| Same exponential base |
\(b^u=b^v\) |
Set \(u=v\). |
| Different exponential bases |
Take \(\ln\) of both sides |
Use the power rule to bring exponents down. |
| Logarithmic form |
\(\log_b(M)=N\) |
Rewrite as \(b^N=M\). |
| Change of base |
\(\log_b(x)=\frac{\ln(x)}{\ln(b)}\) |
Evaluate logs with any base. |
| Log domain |
\(M>0\) |
The logarithm argument must be positive. |
| Graph method |
\(L(x)=R(x)\) |
Solutions are intersections. |
10. Final check
After solving, always check each candidate in the original equation.
- If both sides are defined and equal, the solution is accepted.
- If a logarithm argument is not positive, the solution is rejected.
- If the two sides are not equal, the solution is extraneous.
