Unit 7 Homework 2 Solving Exponential Equations

##Introduction
Solving exponential equations is a central skill assessed in unit 7 homework 2, where students learn to manipulate expressions of the form a^x = b and isolate the unknown exponent. This article provides a clear, step‑by‑step guide, explains the underlying mathematical reasoning, and offers practice strategies that align with typical classroom expectations. By following the outlined methods, learners can confidently tackle homework problems, verify their answers, and deepen their conceptual understanding of exponential functions.

Understanding Exponential Equations

Definition and Basic Form

An exponential equation features a variable in the exponent, such as 2^x = 8. The general form is a^x = c, where a > 0 and a ≠ 1, and c is a positive constant. Recognizing this structure is the first step toward applying appropriate solution techniques.

Key Properties to Remember

• Same Base Property: If a^m = aⁿ and a > 0, a ≠ 1, then m = n.
• Logarithmic Inverse: The logarithm function log_a is the inverse of a^x. Applying a logarithm to both sides can “bring down” the exponent.
• Domain Restrictions: Since exponential expressions are always positive, any solution that would require a negative or zero right‑hand side is invalid.

Step‑by‑Step Approach to Solving

Step 1: Identify the Base and Isolate the Exponential Term

Begin by rewriting the equation so that the exponential expression stands alone on one side. For example, in 5^x‑2 = 25, the exponential term is already isolated.

Step 2: Check for a Common Base

If both sides can be expressed with the same base, rewrite them accordingly. In the example above, 25 = 5², so the equation becomes 5^x‑2 = 5². By the Same Base Property, set the exponents equal: x‑2 = 2, giving x = 4.

Step 3: Apply Logarithms When Bases Differ

When the bases cannot be matched, take the logarithm (common or natural) of both sides. For instance, solve 3^x = 7 by applying log to each side:

[ \log(3^x) = \log 7 ;\Longrightarrow; x \log 3 = \log 7 ;\Longrightarrow; x = \frac{\log 7}{\log 3} ]

Step 4: Solve for the Variable

After isolating x, perform the necessary arithmetic. In the previous example, x ≈ 1.7712 when using common logarithms.

Step 5: Verify the Solution

Substitute the found value back into the original equation to confirm equality. Verification helps catch algebraic slip‑ups and reinforces confidence in the method.

Common Techniques and Examples ### Technique 1: Matching Bases

Example: Solve 4^x+1 = 64.

• Rewrite 64 as 4³ (since 4³ = 64).
• Equation becomes 4^x+1 = 4³.
• Set exponents equal: x + 1 = 3 → x = 2.

Technique 2: Using Natural Logarithms

Example: Solve e^2x = 15.

• Take the natural log of both sides: ln(e^2x) = ln 15.
• Simplify using ln(e^y) = y: 2x = ln 15.
• Solve: x = (ln 15) / 2 ≈ 1.255.

Technique 3: Combining Techniques

Example: Solve 2^x·3^x = 72.

• Combine bases: (2·3)^x = 72 → 6^x = 72.
• Express 72 as 6³ (since 6³ = 216, actually 6² = 36, 6³ = 216; correct approach is to write 72 = 6^log₆72 but easier: take logs).
• Apply log: x log 6 = log 72 → x = log 72 / log 6 ≈ 2.5.

Practice Problems from Unit 7 Homework 2

Below is a concise list of typical problems that appear in unit 7 homework 2. Attempt each using the steps outlined above; the solutions are provided for self‑checking.

1. 5^x‑3 = 125
2. 7^2x = 343
   3

Solutions to the listedproblems

1. (5^{x-3}=125)
   Recognize that (125 = 5^{3}).
   Hence (5^{x-3}=5^{3}) ⇒ (x-3=3) ⇒ (x=6).

2. (7^{2x}=343) Since (343 = 7^{3}), rewrite as (7^{2x}=7^{3}).
   Equate exponents: (2x=3) ⇒ (x=\frac{3}{2}=1.5).

3. (Continuing the pattern) (2^{x+1}=16)
   Write (16) as (2^{4}).
   Then (2^{x+1}=2^{4}) ⇒ (x+1=4) ⇒ (x=3).

Additional practice problems

4. (9^{x}=81)
   Both sides are powers of (3): (9=3^{2}) and (81=3^{4}).
   Thus ((3^{2})^{x}=3^{4}) ⇒ (3^{2x}=3^{4}) ⇒ (2x=4) ⇒ (x=2).

5. (e^{3x}=20)
   Apply the natural logarithm: (\ln(e^{3x})=\ln20) ⇒ (3x=\ln20) ⇒ (x=\frac{\ln20}{3}\approx0.998).

6. (4^{x}\cdot 2^{x}=32)
   Combine the bases: ((4\cdot2)^{x}=8^{x}=32).
   Since (32=2^{5}) and (8=2^{3}), rewrite as ((2^{3})^{x}=2^{5}) ⇒ (2^{3x}=2^{5}) ⇒ (3x=5) ⇒ (x=\frac{5}{3}\approx1.667).

Tips for efficient solving

• Look for hidden powers: Numbers like 27, 64, 125, 256 are often perfect powers of small integers; spotting them lets you match bases without logs.
• Use the change‑of‑base formula when you need a decimal approximation: (\displaystyle \frac{\log a}{\log b} = \frac{\ln a}{\ln b}).
• Check for extraneous results only when you manipulate the equation in ways that could introduce domain issues (e.g., squaring both sides). Pure exponential/logarithmic steps preserve equivalence, so verification is mainly a safeguard against arithmetic slips.

Conclusion

Mastering exponential equations hinges on two core strategies: rewriting each side with a common base whenever possible, and resorting to logarithms when the bases differ. By systematically isolating the exponential term, applying the appropriate property, and verifying the solution, you can tackle a wide range of problems—from simple integer‑exponent cases to those requiring natural or common logs. Consistent practice with the outlined steps will build confidence and speed, ensuring you can solve exponential equations accurately and efficiently.

Extending the Toolbox: MoreComplex Scenarios

When the unknown appears in more than one exponent, or when the equation mixes exponential terms with polynomial expressions, the same foundational ideas still apply—but they often require an extra layer of manipulation.

1. Equations with Two Exponential Terms

Consider

[ 2^{x}=3^{x-1}. ]

Taking logarithms of both sides yields

[ x\ln2=(x-1)\ln3. ]

Now isolate (x):

[ x\ln2=x\ln3-\ln3\quad\Longrightarrow\quad x(\ln2-\ln3)=-\ln3\quad\Longrightarrow\quad x=\frac{-\ln3}{\ln2-\ln3}. ]

Because (\ln2-\ln3) is negative, the fraction simplifies to a positive value. This technique—applying a logarithm, expanding, then collecting the unknown factor—works for any equation of the form (a^{x}=b^{x+c}).

2. Substitution When Bases Differ but Share a Common Factor Suppose

[ (2^{3})^{x}=5^{x+2}. ]

Rewrite the left‑hand side as (2^{3x}). The equation becomes

[ 2^{3x}=5^{x+2}. ]

Now take logarithms:

[ 3x\ln2=(x+2)\ln5. ]

Expand and solve for (x):

[ 3x\ln2=x\ln5+2\ln5\quad\Longrightarrow\quad x(3\ln2-\ln5)=2\ln5\quad\Longrightarrow\quad x=\frac{2\ln5}{3\ln2-\ln5}. ]

If the bases are not directly related, substitution can still reduce the problem to a linear equation in (x) after taking logs.

3. Exponential‑Polynomial Blends

Equations such as

[3^{x}=x+5 ]

cannot be solved algebraically with elementary functions; they require numerical methods (e.g., Newton‑Raphson) or graphing utilities. The process, however, remains conceptually simple:

1. Plot (y=3^{x}) and (y=x+5) on the same axes.
2. Identify the intersection point(s) using a calculator or software.
3. Verify the solution by substitution.

Understanding that some problems shift from closed‑form to approximation equips learners for real‑world modeling where exact analytic solutions are rare.

4. Real‑World Contexts

• Population growth: A species doubles every 7 years. If the initial count is 1,200, the population after (t) years is (1200\cdot2^{t/7}). To find when the population reaches 10,000, set (1200\cdot2^{t/7}=10{,}000) and solve for (t) using logarithms.
• Radioactive decay: The remaining mass after (t) days is (M_0\left(\frac12\right)^{t/10}). Solving for the half‑life when a certain fraction remains involves the same logarithmic steps. * Finance: Compound interest formulas (A=P(1+r)^t) are solved for the time (t) when a target amount (A) is reached, again employing logarithms to isolate the exponent.

These applications illustrate why fluency in manipulating exponential equations is more than an academic exercise—it is a practical skill for interpreting growth, decay, and scaling phenomena.

5. Common Pitfalls and How to Avoid Them