🎯⭐ INTERACTIVE LESSON

Solving Exponential Equations

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Solving Exponential Equations - Complete Interactive Lesson

Part 1: What Makes an Equation "Exponential"?

📈 Solving Exponential Equations

Part 1 of 5 — What Makes an Equation "Exponential"?

Topics in This Part

Section
The Variable Is in the Exponent
Exponent Rules You'll Reuse
Spotting a Common Base

🔑 Key Concept: In an exponential equation, the unknown lives in the exponent — like $2^x = 8$ — not in the base. That one difference changes everything about how we solve.

The Variable Is in the Exponent

Compare these two equations:

Equation	Type	Unknown is...
$x^2 = 16$	quadratic	the base
$2^x = 16$

Concept Check 🎯

Exponent Rules You'll Reuse

Solving these equations leans on the laws of exponents. Keep these close:

Rule	Formula	Example
Product	$b^m \cdot b^n = b^{m+n}$

Rewrite as a Power of the Same Base 🧮

Every number below is a power of a single small base. Enter the exponent.

1) $64 = 2^{\,?}$ 2) $81 = 3^{\,?}$

Spotting a Common Base

The cleanest way to solve an exponential equation is to make both sides have the same base. Then the exponents must match.

To do that, you have to recognize when different-looking numbers share a base:

$8 = 2^3, \qquad 4 = 2^2, \qquad 16 = 2^4, \qquad 32 = 2^5$

Find the Common Base 🔽

For each pair of numbers, choose the smallest base that makes both a whole-number power.

Part 2: The Same-Base Method

📈 Solving Exponential Equations

Part 2 of 5 — The Same-Base Method

🔑 The One-Line Rule: If $b^M = b^N$ (same base $b$ , with $b > 0$ and ), then . Match the bases, then set the exponents equal.

Part 3: Logarithms: The Tool for Every Other Case

📈 Solving Exponential Equations

Part 3 of 5 — Logarithms: The Tool for Every Other Case

🔑 Why we need logs: $2^x = 10$ has no clean common base — $10$ isn't a power of $2$ . The logarithm is the operation that finally answers " $2$ to what power gives ?"

Part 4: Solving with Logs & Real Applications

📈 Solving Exponential Equations

Part 4 of 5 — Solving with Logs & Real Applications

🔑 The Move: When bases won't match, take the log (or $\ln$ ) of both sides, then use $\ln(b^x) = x \ln b$ to bring the exponent down where you can solve for it.

Taking the Log of Both Sides

The power rule of logs, , is what frees the exponent.

Part 5: Mixed Practice & Mastery Check

📈 Solving Exponential Equations

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) match common bases, (2) read a log as an exponent, (3) change base with $\ln$ , and (4) solve real growth/decay problems. Time to mix it all together — plus one new twist.

A New Twist: Quadratic in Form

Some equations hide a quadratic inside. Watch for $b^{2x}$ , which equals $(b^x)^2$ .

For $x^2 = 16$ you take a square root. But for $2^x = 16$ , taking a root does nothing useful — you need a different idea.

The big question of this whole lesson is:

🔑 Core Question: "To what power must I raise the base to get this number?"

For $2^x = 16$ , the answer is $x = 4$ , because $2^4 = 16$ . The rest of the lesson is about answering that question even when it isn't obvious.

💡 The power of a power rule is the workhorse. It's what lets us rewrite something like $9^x$ as $(3^2)^x = 3^{2x}$ — turning two different bases into one.

So $8$ , $4$ , $16$ , and $32$ are all powers of $2$ . Likewise $9 = 3^2$ and $27 = 3^3$ are both powers of $3$ .

🔑 Preview: If you can write an equation as $b^{(\text{stuff})} = b^{(\text{other stuff})}$ with the same $b$ , then stuff = other stuff. That single move — coming in Part 2 — solves a huge family of problems.

The Method

Step 1. Rewrite both sides as powers of the same base. Step 2. Set the exponents equal. Step 3. Solve the resulting (usually linear) equation.

Worked Example: $2^{x+1} = 32$

Write $32$ as a power of $2$ : $\;32 = 2^5$ .

$2^{x+1} = 2^5 \;\Rightarrow\; x + 1 = 5 \;\Rightarrow\; x = 4$

✅ Check: $2^{4+1} = 2^5 = 32$ ✓

Concept Check 🎯

When the Bases Look Different

Sometimes the two bases aren't equal yet — but both are powers of a smaller base. Use $(b^m)^n = b^{mn}$ to fix that.

Worked Example: $9^x = 27$

Both sides are powers of $3$ : $\;9 = 3^2$ and $27 = 3^3$ .

$\left(3^2\right)^x = 3^3 \;\Rightarrow\; 3^{2x} = 3^3 \;\Rightarrow\; 2x = 3 \;\Rightarrow\; x = \tfrac{3}{2}$

Worked Example: $8^x = 4^{x+1}$

Write everything as a power of $2$ : $\;8 = 2^3$ , $\;4 = 2^2$ .

$\left(2^3\right)^x = \left(2^2\right)^{x+1} \;\Rightarrow\; 2^{3x} = 2^{2x+2} \;\Rightarrow\; 3x = 2x + 2 \;\Rightarrow\; x = 2$

⚠️ Watch the parentheses: $\left(2^2\right)^{x+1}$ means the entire exponent $x+1$ gets multiplied by , giving — not .

Walk the Steps 🔽

You're solving $4^{x-1} = 64$ . Choose what happens at each stage.

Same-Base Checklist

Before you reach for a calculator, run through this:

Ask	If yes...
Are both sides already the same base?	Set exponents equal.
Are they powers of a smaller base?	Rewrite with $(b^m)^n = b^{mn}$ , then match.
Is a side a single number like $81$ ?	Express it as a power: $81 = 3^4$ .

💡 The same-base method gives exact answers with no rounding. Always try it first; only switch to logs (Part 3) when no common base exists.

Solve by Matching Bases 🧮

Rewrite each side with a common base, then solve for $x$ . (Decimals or fractions are fine.)

1) $3^{x} = 81 \;\Rightarrow\; x = \,?$ 2) $2^{3x} = 32 \;\Rightarrow\; x = \,?$ 3) $25^{x} = 125 \;\Rightarrow\; x = \,?$

What a Logarithm Is

A logarithm is just an exponent in disguise. The statement

$\log_b N = x \qquad \text{means exactly} \qquad b^x = N$

Read $\log_b N$ as "the power you raise $b$ to in order to get $N$ ."

Logarithm form	Exponential form	Value
$\log_2 8$	$2^x = 8$

🔑 The two-way switch: $b^x = N \;\Longleftrightarrow\; \log_b N = x$ . Logs and exponentials undo each other — they are inverse operations.

Concept Check 🎯

Two Logs Your Calculator Has

Most calculators only have two log buttons, so we lean on them:

Common log $\log x$ — base $10$ (the " $\log$ " key).
Natural log $\ln x$ — base $e$ , where $e \approx 2.718$ (the " $\ln$ " key).

These two are enough to solve anything, thanks to the change-of-base formula:

$\log_b N = \frac{\ln N}{\ln b} = \frac{\log N}{\log b}$

Example: Evaluate $\log_5 20$

$\log_5 20 = \frac{\ln 20}{\ln 5} \approx \frac{2.996}{1.609} \approx 1.861$

💡 You can divide by either $\ln$ or $\log$ — just be consistent on top and bottom. Both give the same answer.

Switch Between Forms 🔽

Match each exponential statement to its logarithmic twin (and vice-versa).

Using Change of Base in Practice

Your calculator can't compute $\log_3 50$ directly, but change of base fixes that:

$\log_3 50 = \frac{\ln 50}{\ln 3} = \frac{3.912}{1.099} \approx 3.56$

✅ Sanity check: $3^{3.56} \approx 50$ , so the answer is reasonable — $\log_3 50$ should land between $3$ (since ) and (since ), and does.

Change of Base 🧮

Use $\log_b N = \dfrac{\ln N}{\ln b}$ . Round to two decimal places.

1) $\log_3 50 = \,?$ 2) $\log_2 10 = \,?$

\log(b^x) = x \log b

Worked Example: $5^x = 20$

$\ln(5^x) = \ln 20 \;\Rightarrow\; x \ln 5 = \ln 20 \;\Rightarrow\; x = \frac{\ln 20}{\ln 5} \approx 1.86$

Worked Example: $e^{x} = 7$

Because the base is $e$ , use $\ln$ — and $\ln(e^x) = x$ :

$\ln(e^x) = \ln 7 \;\Rightarrow\; x = \ln 7 \approx 1.95$

💡 Rule of thumb: base $e$ → use $\ln$ ; base $10$ → use $\log$ ; any other base → use either, with change-of-base.

Bring the Exponent Down 🔽

You're solving $3^x = 50$ by taking $\ln$ of both sides. Choose each step.

Isolate the Power First

If something is multiplied onto (or added to) the exponential, undo that before taking a log.

Worked Example: $4 \cdot 3^{x} = 36$

Divide by $4$ first: $\;3^x = 9$ . Now it's a same-base problem!

$3^x = 3^2 \;\Rightarrow\; x = 2$

Worked Example: $e^{3x} = 20$

The power is already alone, so take $\ln$ :

$3x = \ln 20 \;\Rightarrow\; x = \frac{\ln 20}{3} \approx 1.00$

⚠️ Order matters. In $4 \cdot 3^x = 36$ you must divide by $4$ first. Taking a log of $4 \cdot 3^x$ before isolating creates a mess () — correct, but far harder.

Concept Check 🎯

Where This Shows Up: Growth & Decay

Exponential equations model money, populations, and radioactive decay. Solving them means finding a time.

Compound Interest: How long to double $1000 at 5%?

The balance is $A = 1000(1.05)^t$ . Set $A = 2000$ :

$1000(1.05)^t = 2000 \;\Rightarrow\; 1.05^t = 2 \;\Rightarrow\; t = \frac{\ln 2}{\ln 1.05} \approx 14.2 \text{ years}$

Half-Life Decay: $80$ mg, halving every $6$ hours, down to $10$ mg?

$80\left(\tfrac{1}{2}\right)^{t/6} = 10 \;\Rightarrow\; \left(\tfrac{1}{2}\right)^{t/6} = \tfrac{1}{8} = \left(\tfrac{1}{2}\right)^3 \;\Rightarrow\; \tfrac{t}{6} = 3 \;\Rightarrow\; t = 18 \text{ hours}$

💡 Notice the decay problem reduced to a same-base equation — always check whether the numbers line up before reaching for a calculator log.

Application Practice 🧮

1) Solve $2^x = 10$ for $x$ , rounded to two decimals. 2) A $500 investment grows as $500(1.04)^t$ . To reach $2000 you need $1.04^t = 4$ . Find $t = \dfrac{\ln 4}{\ln 1.04}$ , rounded to the nearest whole year. 3) Solve $500 \cdot 2^{\,t/10} = 4000$ exactly (it's a same-base problem): $t = \,?$

Worked Example: $2^{2x} - 5 \cdot 2^{x} + 4 = 0$

Let $u = 2^x$ . Then $2^{2x} = (2^x)^2 = u^2$ :

$u^2 - 5u + 4 = 0 \;\Rightarrow\; (u - 1)(u - 4) = 0 \;\Rightarrow\; u = 1 \text{ or } u = 4$

Now switch back to $x$ using $u = 2^x$ :

$2^x = 1 \Rightarrow x = 0, \qquad 2^x = 4 \Rightarrow x = 2$

⚠️ Don't stop at $u$ . The variable is $x$ , not $u$ . You must replace $u$ with $2^x$ and solve for $x$ at the end. Also: if a $u$ -value is negative or zero, discard it — $b^x$ is always positive.

Mixed Practice 🎯

Quick Reference

Situation	Strategy
Both sides share a base (or can)	Match bases, set exponents equal
Bases can't match	Take $\ln$ / $\log$ of both sides
Base is $e$ / base is $10$	Use $\ln$ / use $\log$
Need $\log_b N$ on a calculator	$\dfrac{\ln N}{\ln b}$ (change of base)
Coefficient on the power	Isolate the power first
$b^{2x}$ appears	Substitute $u = b^x$ (quadratic in form)

🔑 The golden order: isolate the power → match bases if you can → otherwise take a log → solve → and check by plugging back in.

Mixed Solve 🧮

1) $6^x = 216 \;\Rightarrow\; x = \,?$ (same base — exact answer) 2) $e^{x} = 0.5 \;\Rightarrow\; x = \ln(0.5) = \,?$ (round to two decimals) 3) From $2^{2x} - 5\cdot 2^x + 4 = 0$ , the two solutions are $x = 0$ and $x = \,?$

You're Ready

You started by asking "to what power?" and now you have a full toolkit:

Same base → set exponents equal (exact, no rounding).
No common base → take $\ln$ or $\log$ and use $\ln(b^x) = x\ln b$ .
$\log_b N$ on a calculator → change of base, $\dfrac{\ln N}{\ln b}$ .
$b^{2x}$ pattern → substitute $u = b^x$ and factor.

🔑 One last reminder: $b^x$ is always positive. If an answer would require $b^x \le 0$ , that branch has no real solution. The Exit Quiz below confirms you've got it.

Exit Quiz ✅

Answer all three to finish the lesson.

Solving Exponential Equations

Solving Exponential Equations - Complete Interactive Lesson

Part 1: What Makes an Equation "Exponential"?

📈 Solving Exponential Equations

Topics in This Part

The Variable Is in the Exponent

Exponent Rules You'll Reuse

Spotting a Common Base

Part 2: The Same-Base Method

📈 Solving Exponential Equations

Part 3: Logarithms: The Tool for Every Other Case

📈 Solving Exponential Equations

Part 4: Solving with Logs & Real Applications

📈 Solving Exponential Equations

Taking the Log of Both Sides

Part 5: Mixed Practice & Mastery Check

📈 Solving Exponential Equations

A New Twist: Quadratic in Form

The Method

Worked Example: 2x+1=322^{x+1} = 322x+1=32

When the Bases Look Different

Worked Example: 9x=279^x = 279x=27

Worked Example: 8x=4x+18^x = 4^{x+1}8x=4x+1

Same-Base Checklist

What a Logarithm Is

Two Logs Your Calculator Has

Example: Evaluate log⁡520\log_5 20log5​20

Using Change of Base in Practice

Worked Example: 5x=205^x = 205x=20

Worked Example: ex=7e^{x} = 7ex=7

Isolate the Power First

Worked Example: 4⋅3x=364 \cdot 3^{x} = 364⋅3x=36

Worked Example: e3x=20e^{3x} = 20e3x=20

Where This Shows Up: Growth & Decay

Compound Interest: How long to double $1000 at 5%?

Half-Life Decay: 808080 mg, halving every 666 hours, down to 101010 mg?

Worked Example: 22x−5⋅2x+4=02^{2x} - 5 \cdot 2^{x} + 4 = 022x−5⋅2x+4=0

Quick Reference

You're Ready

Worked Example: $2^{x+1} = 32$

Worked Example: $9^x = 27$

Worked Example: $8^x = 4^{x+1}$

Example: Evaluate $\log_5 20$

Worked Example: $5^x = 20$

Worked Example: $e^{x} = 7$

Worked Example: $4 \cdot 3^{x} = 36$

Worked Example: $e^{3x} = 20$

Half-Life Decay: $80$ mg, halving every $6$ hours, down to $10$ mg?

Worked Example: $2^{2x} - 5 \cdot 2^{x} + 4 = 0$