Solving Exponential and Logarithmic Equations Exactly

1. Solving Exponential Equations

An exponential equation is one where the variable you need to solve for is in the exponent, like $2^x=32$ or $5^{t-1}=10$. There are two primary algebraic strategies for solving these, depending on the structure of the equation.

Strategy 1: Using a Common Base

The simplest case is when you can express both sides of the equation as powers of the same base. This relies on the one-to-one property of exponential functions: if the bases are equal, the exponents must also be equal.

If $b^M=b^N$, then $M=N$ (where $b>0$ and $b\ne 1$).

For example, to solve $2^x=32$, we can rewrite 32 as a power of 2:
$2^x=2^5$
Because the bases are both 2, we can equate the exponents:
$x=5$

This method becomes powerful when the expressions are more complex. You'll need to use your knowledge of index laws to manipulate the terms.

The following resource provides a clear guide on this technique. Pay close attention to the examples where both sides of the equation must be rewritten.

To see this method applied in a dynamic way, the following video segment provides a concise walkthrough.

Strategy 2: Taking Logarithms of Both Sides

What happens when you can't find a common base, as in an equation like $3^x=11$? In this situation, our main tool is to take the logarithm of both sides. This is effective because of the logarithm power rule, $\log (A^p)=p\log (A)$, which allows us to bring the variable down from the exponent.

You can use any base for the logarithm, but the most common choices for IB Mathematics are the natural logarithm ($\ln$, base $e$) or the common logarithm ($\log$, base 10), as these are readily available on your calculator. In most cases, the natural logarithm is preferred.

Let's solve $3^x=11$ using this method:

1. Take the natural log of both sides:
   $\ln (3^x)=\ln (11)$
2. Use the power rule to bring the exponent down:
   $x\ln (3)=\ln (11)$
3. Solve for x:
   $x=\frac{\ln (11)}{\ln (3)}$

This is the exact solution. For a Paper 1 question, this is where you would stop. For a Paper 2 question, you could then use your TI-84 to find the decimal approximation.

The following resources provide more detail and examples, including cases with variables on both sides.

For a visual guide on a more complex problem, check out this video clip.

2. Solving Logarithmic Equations

Solving logarithmic equations often involves "undoing" the logarithm by converting the equation into its equivalent exponential form. The key relationship to remember is:

${\log }_b(A)=C⟺b^C=A$

Before we dive into the methods, a critical warning: the argument of a logarithm must be positive. This means that after you find a potential solution, you must check that it doesn't result in taking the logarithm of a zero or a negative number in the original equation. Solutions that are algebraically correct but violate this domain condition are called extraneous solutions and must be rejected.

Strategy 1: Rewriting in Exponential Form

If your equation can be arranged into the form ${\log }_b(\text{expression})=C$, the most direct approach is to convert it to its exponential form.

For example, to solve ${\log }_3(x+5)=2$:

1. Convert to exponential form:
   $3^2=x+5$
2. Solve for x:
   $9=x+5$
   $x=4$
3. Check the solution: Plugging $x=4$ into the original equation gives ${\log }_3(4+5)={\log }_3(9)$. Since the argument (9) is positive, the solution is valid.

The following video provides an excellent series of examples that build in complexity using this fundamental technique.

Strategy 2: Using Properties of Logarithms

Many IB problems involve multiple logarithm terms. Your goal is to use the logarithm laws (product, quotient, power) to condense the log terms into a single logarithm. Once you have an equation in the form ${\log }_b(\text{expression})=C$, you can use Strategy 1.

Logarithm Laws (from the IB Formula Booklet):

• ${\log }_{c}a+{\log }_{c}b={\log }_c(ab)$
• ${\log }_{c}a-{\log }_{c}b={\log }_c(a/b)$
• ${\log }_c{a}^m=m{\log }_{c}a$

Consider the equation: ${\log }_2(x)+{\log }_2(x-2)=3$

1. Condense using the product rule:
   ${\log }_2(x(x-2))=3$

2. Convert to exponential form:
   $2^3=x(x-2)$

3. Solve the resulting (quadratic) equation:
   $8=x^2-2x$
   $x^2-2x-8=0$
   $(x-4)(x+2)=0$
   This gives potential solutions $x=4$ and $x=-2$.

4. Check for extraneous solutions:

   • For $x=4$: The original equation is ${\log }_2(4)+{\log }_2(4-2)={\log }_2(4)+{\log }_2(2)$. Both arguments are positive. This solution is valid.
   • For $x=-2$: The original equation is ${\log }_2(-2)+{\log }_2(-2-2)$. The first term involves the logarithm of a negative number. This solution is extraneous and must be rejected.

The only solution is $x=4$. The video you just reviewed provides excellent examples of this process.

A special case of this is when you can create a single logarithm on both sides of the equation. This uses the one-to-one property of logarithms:

If ${\log }_b(M)={\log }_b(N)$, then $M=N$.

This simplifies the process as you can directly equate the arguments, but you must still check for extraneous solutions at the end.

3. IB-Style Problem: The Disguised Quadratic

A classic HL problem type involves an exponential equation that is actually a quadratic equation in disguise.
Let's solve the equation: $4^{x+1}-9\cdot 2^x+2=0$.

1. Rewrite with a common base. Notice that $4=2^2$.
   $(2^2{)}^{x+1}-9\cdot 2^x+2=0$
   $2^{2(x+1)}-9\cdot 2^x+2=0$
   $2^{2x+2}-9\cdot 2^x+2=0$
   Using index laws, we can write $2^{2x+2}$ as $2^{2x}\cdot 2^2$, which is $4\cdot (2^x{)}^2$.
   $4\cdot (2^x{)}^2-9\cdot (2^x)+2=0$

2. Make a substitution. The equation now has a repeated term, $2^x$. Let $u=2^x$.
   $4u^2-9u+2=0$

3. Solve the quadratic. We can factor this quadratic:
   $(4u-1)(u-2)=0$
   So, $u=\frac{1}{4}$ or $u=2$.

4. Reverse the substitution and solve for x.

   • Case 1: $u=\frac{1}{4}$
     $2^x=\frac{1}{4}$
     $2^x=2^{-2}$
     $x=-2$
   • Case 2: $u=2$
     $2^x=2$
     $2^x=2^1$
     $x=1$

The solutions are $x=-2$ and $x=1$. Since these were from an exponential equation (whose domain is all real numbers), we do not need to check for extraneous solutions in the same way we do for logarithms.

Conclusion

In this lesson, we have developed a powerful set of algebraic tools for solving exponential and logarithmic equations. Your ability to recognize the structure of an equation and choose the right strategy is key.

Key Takeaways:

• Exponential Equations:
  • If possible, rewrite both sides with a common base and equate the exponents.
  • If not, take the logarithm (usually $\ln$) of both sides and use the power rule to solve for the variable.
• Logarithmic Equations:
  • Use logarithm laws to condense multiple log terms into a single one.
  • Rewrite the equation in exponential form to eliminate the logarithm.
  • Always check your final answers in the original equation to discard any extraneous solutions that would lead to the logarithm of a non-positive number.

These skills are not isolated tricks; they are fundamental components of mathematical reasoning that you will apply repeatedly in calculus and when modeling real-world phenomena.

In our next lesson, we will move on to solving inequalities, specifically those involving polynomials and rational expressions. This will introduce the methods of using critical values and sign analysis, further honing the analytical skills needed for success in calculus.