BetterGrades Algebra · Unit A13 · Lesson

Solving exponential equations

Use common bases or logarithms and preserve exact form before decimal approximation.

Opening situation

Start here

Find time, rate, or exponent in a growth model.

Use the opening situation and three distinct, fully solved cases to learn solving exponential equations as a connected mathematical idea rather than a memorized slogan.

Before this lesson

Prerequisite check

  1. State the earlier definition or operation most directly connected to: Use common bases or logarithms and preserve exact form before decimal approximation.
  2. Classify the object in the worked prompt before choosing an operation: Solve 52(3x)=705\cdot 2^(3x) = 70 exactly and then approximate.
  3. Name the check you would use to reject an answer with the wrong sign, domain, units, endpoint, or graph behavior.
Lesson text

Explanation

Use common bases or logarithms and preserve exact form before decimal approximation. The lesson is about a particular mathematical decision, not a keyword or a decorative notation pattern. In solving exponential equations, first identify the object being studied and the information the answer must contain. Then mark the conditions that cannot be lost: these may include sign, endpoint inclusion, grouping, units, denominator restrictions, real-number domain, or the difference between an exact value and an approximation. A useful solution explains why its first move matches that structure.

Find time, rate, or exponent in a growth model. This opening is useful because it forces the quantities to acquire meaning before symbols compress them. Name the changing and fixed quantities, define any reference value or input interval, and decide what would count as a plausible result. An estimate, sign prediction, graph feature, or domain statement made before calculation becomes an independent check afterward. Without that prediction, algebra can be internally tidy while answering the wrong contextual question.

Consider the worked problem: Solve 52(3x)=705\cdot 2^(3x) = 70 exactly and then approximate. Begin with this justified move: Divide by 55 to isolate the exponential expression. Next, take logarithms and use the power law. Finally, solve for xx and postpone decimal evaluation until the exact form is complete. Each line should preserve the relevant relationship or deliberately produce candidates that are later tested. Skipping the middle line may hide the exact sign, factor, interval, or restriction on which the conclusion depends.

The result is x=ln(14)3ln21.269x = \frac{ln(14)}{3ln 2} \approx 1.269. Logarithms invert exponentiation when a convenient common base is unavailable. A textbook answer does not stop at the last symbol. It states what the result means, includes units or set notation where required, and distinguishes a verified solution from a candidate. The original statement remains the final authority whenever the method includes a one-way operation, denominator clearing, squaring, graph estimation, regression, or numerical approximation.

Use a table of differences or ratios, an exponential formula, a graph with asymptote, and the equivalent logarithmic statement. Changing representation is useful only when it exposes information rather than duplicating decoration. A table may reveal constant difference or ratio, a graph may reveal intersections or extrema, interval notation may compress a truth set, and factored or vertex form may expose a feature hidden in expanded form. The second representation must preserve the same values, restrictions, units, endpoints, and conclusions as the first.

Linear change adds a constant amount over equal input intervals; exponential change multiplies by a constant factor. In aa table, constant differences signal linear structure and constant ratios signal exponential structure. A repeated percent change uses the multiplier 1+r1 + r for growth or 1r1 - r for decay, so equal percentages compound rather than add. For solving exponential equations, connect this principle directly to the stated outcome: Use common bases or logarithms and preserve exact form before decimal approximation.

An exponential function f(x)=f(x) = abˣ has initial value a and base b, with bb positive and not equal to one. The base determines growth or decay, while transformations shift, scale, or reflect the graph and move its horizontal asymptote. Models require a meaningful time unit and domain. Compound interest distinguishes nominal rate from the rate per compounding period, and continuous change uses ee as the natural limiting base. For solving exponential equations, connect this principle directly to the stated outcome: Use common bases or logarithms and preserve exact form before decimal approximation.

A logarithm answers an exponent question. The statement log_b(y) =x= x is equivalent to bˣ == y, with b>0,b1,b > 0, b \ne 1, and y>0y > 0. Logarithm laws follow from exponent laws: products become sums, quotients become differences, and powers become coefficients. There is no corresponding rule that splits log(a ++ b). Solving logarithmic equations requires every final log argument to remain positive. For solving exponential equations, connect this principle directly to the stated outcome: Use common bases or logarithms and preserve exact form before decimal approximation.

A common failure is: Adding a percent repeatedly, treating a logarithm as an ordinary factor, or applying a false sum law. Exponential change compounds multiplicatively and logarithm laws translate exponent structure, not arbitrary addition. The repair is concrete: Write the multiplier or equivalent exponential equation, preserve base and argument restrictions, and check the result in the original model. In the worked case, use the repair by checking “x=ln(14)3ln21.269x = \frac{ln(14)}{3ln 2} \approx 1.269.” against the original problem rather than trusting that the final line merely looks familiar.

Logarithms invert exponentiation when a convenient common base is unavailable. That conclusion is the bridge to the next lesson: the method matters because it preserves meaning while the representation changes. A durable summary therefore has four parts—classify the object, state the conditions, carry out one justified step at aa time, and perform an independent check. If any of those parts is missing, return to the original quantities before adding more algebra.

Method

Solve solving exponential equations from structure

  1. Divide by 55 to isolate the exponential expression.
  2. Take logarithms and use the power law.
  3. Solve for xx and postpone decimal evaluation until the exact form is complete.

Check: Verify the initial value, per-period multiplier, domain, and any logarithmic candidate in the original exponential or log equation.

Reference

Definitions and conditions

Solving exponential equations
Use common bases or logarithms and preserve exact form before decimal approximation.Use the term only when the object satisfies the structural and domain conditions developed in this lesson.
growth factor
The constant multiplier applied during each equal input interval.For percent rate r, the factor is 1+r1 + r for growth and 1r1 - r for decay.
logarithm
The exponent to which a valid base must be raised to produce a positive argument.The base is positive and not one; the argument is positive.
horizontal asymptote
A horizontal line approached by a function’s outputs as inputs move in a direction.A model may approach the line without reaching it in its theoretical domain.
Examples

Worked examples

Worked Example 1

Solve 52(3x)=705\cdot 2^(3x) = 70 exactly and then approximate.

  1. Divide by 55 to isolate the exponential expression.
  2. Take logarithms and use the power law.
  3. Solve for xx and postpone decimal evaluation until the exact form is complete.

Answerx=ln(14)3ln21.269x = \frac{ln(14)}{3ln 2} \approx 1.269

Logarithms invert exponentiation when a convenient common base is unavailable.

Worked Example 2

Solve 73(2x1)=507\cdot 3^(2x-1) = 50 exactly and approximately.

  1. Divide by 77.
  2. Take logarithms: (2x1)ln3=ln(507)(2x - 1)ln3 = ln(\frac{50}{7}).
  3. Solve forxx

Answerx=1+ln(507)ln321.395x = \frac{1 + \frac{ln(\frac{50}{7})}{ln}3}{2} \approx 1.395

Logarithms isolate a variable that appears in an exponent.

Worked Example 3

Solve 4(x+1)=8(2x1)4^(x+1) = 8^(2x-1) by using a common base.

  1. Rewrite4=228=234 = 2^{2} \qquad 8 = 2^{3}
  2. Equate exponents: 2(x+1)=3(2x1)2(x + 1) = 3(2x - 1).
  3. Solve the linear equation.

Answerx=54x = \frac{5}{4}

A common base avoids decimal logarithms and preserves an exact rational result.

Practice

20 practice questions

Recall and read the structure

Warm-up

Question 1Retrieval · Foundation

Classify the mathematical object and requested action in this lesson case: Solve 52(3x)=705\cdot 2^(3x) = 70 exactly and then approximate.

Need a hint?

Recall the named definition or perform a direct substitution before choosing an operation.

Question 2Retrieval · Foundation

State the central definition behind this outcome: Use common bases or logarithms and preserve exact form before decimal approximation.

Need a hint?

Recall the named definition or perform a direct substitution before choosing an operation.

Question 3Concept · Standard

Before calculating, list every sign, endpoint, unit, grouping, or domain condition that can affect: Solve 52(3x)=705\cdot 2^(3x) = 70 exactly and then approximate.

Need a hint?

State what must remain true, then connect that condition to the equation.

Question 4Concept · Standard

Explain why this opening move is valid: Divide by 55 to isolate the exponential expression.

Need a hint?

State what must remain true, then connect that condition to the equation.

Build accuracy one step at a time

Core practice

Question 5Procedure · Standard

Solve 52(3x)=705\cdot 2^(3x) = 70 exactly and then approximate.

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 6Procedure · Standard

Solve 73(2x1)=507\cdot 3^(2x-1) = 50 exactly and approximately.

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 7Procedure · Mixed

Solve 4(x+1)=8(2x1)4^(x+1) = 8^(2x-1) by using a common base.

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 8Procedure · Mixed

Verify the proposed result “x=ln(14)3ln21.269x = \frac{ln(14)}{3ln 2} \approx 1.269.” against the original statement.

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 9Procedure · Standard

Complete the calculation after “Divide by 77.” in this problem: Solve 73(2x1)=507\cdot 3^(2x-1) = 50 exactly and approximately.

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 10Procedure · Mixed

Name and justify the most efficient first move, then solve: Solve 4(x+1)=8(2x1)4^(x+1) = 8^(2x-1) by using a common base.

Need a hint?

Write one equality-preserving step at a time and keep signs and grouping visible.

Question 11Representation · Mixed

Compare the methods used in these two cases and identify the structural reason they differ: Solve 73(2x1)=507\cdot 3^(2x-1) = 50 exactly and approximately. Solve 4(x+1)=8(2x1)4^(x+1) = 8^(2x-1) by using a common base.

Need a hint?

Label the quantities and make the same relationship visible in the new form.

Question 12Representation · Transfer

Create the representation most useful for checking this result: Solve 73(2x1)=507\cdot 3^(2x-1) = 50 exactly and approximately. Use a table of differences or ratios, an exponential formula, a graph with asymptote, and the equivalent logarithmic statement.

Need a hint?

Label the quantities and make the same relationship visible in the new form.

Explain, compare, and diagnose

Represent and reason

Question 13Error Analysis · Mixed

A learner reports “x=ln(14)3ln21.269x = \frac{ln(14)}{3ln 2} \approx 1.269.” but omits the original-condition check. Explain the risk before deciding whether the result is supported.

Need a hint?

Locate the first line that no longer preserves the original relationship.

Question 14Transfer · Transfer

Repair a solution that skips “Equate exponents: 2(x+1)=3(2x1)2(x + 1) = 3(2x - 1).” while solving: Solve 4(x+1)=8(2x1)4^(x+1) = 8^(2x-1) by using a common base.

Need a hint?

Identify the familiar equation structure before changing any symbols.

Question 15Modeling · Transfer

In this solving exponential equations case, change one numerical value, solve the revised problem, and identify which parts of the original method still apply: Solve 52(3x)=705\cdot 2^(3x) = 70 exactly and then approximate.

Need a hint?

Define the unknown and its units before writing the equation.

Question 16Exit · Standard

Connect the opening situation “Find time, rate, or exponent in a growth model.” to the algebraic structure used in the worked case. Define quantities and units before writing any equation.

Need a hint?

Solve, classify the solution set, and verify against the original equation.

Model, transfer, and verify

Finish strong

Question 17Transfer · Transfer

Explain why the method for solving exponential equations is valid here and name one nearby problem where it would not apply.

Need a hint?

Identify the familiar equation structure before changing any symbols.

Question 18Modeling · Transfer

Compare the conclusions of all three worked cases with this lesson outcome—Use common bases or logarithms and preserve exact form before decimal approximation. Explain what remains invariant across them.

Need a hint?

Define the unknown and its units before writing the equation.

Question 19Error Analysis · Mixed

Exit check: solve and verify without referring to the displayed steps. Solve 73(2x1)=507\cdot 3^(2x-1) = 50 exactly and approximately.

Need a hint?

Locate the first line that no longer preserves the original relationship.

Question 20Exit · Standard

Exit check: solve and verify without referring to the displayed steps. Solve 4(x+1)=8(2x1)4^(x+1) = 8^(2x-1) by using a common base.

Need a hint?

Solve, classify the solution set, and verify against the original equation.

Common mistakes

Error analysis

Wrong move: Adding a percent repeatedly, treating a logarithm as an ordinary factor, or applying a false sum law.

Why it fails: Exponential change compounds multiplicatively and logarithm laws translate exponent structure, not arbitrary addition.

Repair: Write the multiplier or equivalent exponential equation, preserve base and argument restrictions, and check the result in the original model.

Open-response checkA13.10

Exit check: solve and verify without referring to the displayed steps. Solve 4(x+1)=8(2x1)4^(x+1) = 8^(2x-1) by using a common base.

Write a complete attempt before opening the response guide.

Attempt once to unlock the response guide

Complete a substantive attempt to unlock the protected solution and scoring criteria.

Before continuing

Exit check

  1. Exit check: solve and verify without referring to the displayed steps. Solve 73(2x1)=507\cdot 3^(2x-1) = 50 exactly and approximately.
  2. Exit check: solve and verify without referring to the displayed steps. Solve 4(x+1)=8(2x1)4^(x+1) = 8^(2x-1) by using a common base.
Summary

What to remember

Use common bases or logarithms and preserve exact form before decimal approximation. Use structure to choose the method, preserve every condition, and interpret the checked result.

  • Verify the initial value, per-period multiplier, domain, and any logarithmic candidate in the original exponential or log equation.
  • Logarithms invert exponentiation when a convenient common base is unavailable.

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