Solving Exponential Equations

What it means to “solve” an exponential equation

In an exponential equation, the unknown appears in an exponent, such as 3^{x}=81 or 120(0.92)^{t}=50. The goal is to isolate the variable. Two main pathways are common:

• Same-base strategy: rewrite both sides with a common base, then equate exponents.
• Logarithm strategy: isolate the exponential expression, take a logarithm of both sides, then solve for the variable.

1) Same-base strategy: rewrite with a common base and equate exponents

This method is fastest when both sides can be expressed as powers of the same positive base (not 1). The key move is: if a^{u}=a^{v} with a>0 and a≠1, then u=v.

Example 1: direct common base

Solve 2^{x}=32.

1. Rewrite 32 as a power of 2: 32=2^{5}.
2. Now 2^{x}=2^{5}, so equate exponents: x=5.

Example 2: rewrite both sides

Solve 9^{x}=27.

1. Rewrite with base 3: 9=(3^{2}) and 27=3^{3}.
2. Then 9^{x}=(3^{2})^{x}=3^{2x}.
3. So 3^{2x}=3^{3} ⇒ 2x=3 ⇒ x=3/2.

Example 3: common base after algebraic rearrangement

Solve 5^{2x-1}=125.

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1. Rewrite 125 as 5^{3}.
2. Equate exponents: 2x-1=3.
3. Solve: 2x=4 ⇒ x=2.

Example 4: two exponentials with the same base

Solve 4^{x+1}=4^{2x-3}.

1. Bases already match and are valid (4>0, 4≠1).
2. Equate exponents: x+1=2x-3.
3. Solve: 1=x-3 ⇒ x=4.

When the same-base strategy is not ideal

If rewriting to a common base is messy or impossible with familiar integer bases (e.g., 7^{x}=20 or 3^{x}=2^{x+1}), use logarithms.

2) Cases requiring logarithms: isolate, log both sides, solve

When you cannot conveniently express both sides as powers of the same base, isolate the exponential part and apply a logarithm to both sides. The typical workflow:

• Step A: Isolate the exponential expression (the part with the variable in the exponent).
• Step B: Take log (base 10) or ln (base e) or any convenient base on both sides.
• Step C: Bring the exponent down using the log rule log(a^{k})=k log(a).
• Step D: Solve the resulting linear equation for the variable.

Example 5: simple log solve

Solve 7^{x}=20.

1. Isolate the exponential term: it already is.
2. Take logs: log(7^{x})=log(20).
3. Bring down the exponent: x log(7)=log(20).
4. Solve: x=log(20)/log(7) (approximately 1.537).

Example 6: isolate first (growth/decay style equation)

Solve 120(0.92)^{t}=50 for t.

1. Isolate the exponential: divide by 120: (0.92)^{t}=50/120=5/12.
2. Take logs: ln((0.92)^{t})=ln(5/12).
3. Bring down exponent: t ln(0.92)=ln(5/12).
4. Solve: t=ln(5/12)/ln(0.92) (a positive number because both numerator and denominator are negative).

Example 7: exponential on both sides with different bases

Solve 3^{x}=2^{x+1}.

1. Take logs on both sides: ln(3^{x})=ln(2^{x+1}).
2. Bring down exponents: x ln 3=(x+1) ln 2.
3. Expand and solve: x ln 3=x ln 2+ln 2 ⇒ x(ln 3- ln 2)=ln 2.
4. x=ln 2/(ln 3- ln 2).

Example 8: isolate a shifted exponential

Solve 2^{x}+5=17.

1. Isolate the exponential: 2^{x}=12.
2. Take logs: ln(2^{x})=ln(12).
3. Bring down exponent: x ln 2=ln 12.
4. Solve: x=ln 12/ln 2 (which equals log_{2}(12)).

3) Choosing log, ln, or any base: what changes and what doesn’t

You can use any logarithm base as long as you use it consistently on both sides. The solution for the variable will be the same number, even if the expression looks different.

Base 10 (log) vs base e (ln)

• Use ln often in continuous growth/decay contexts and many scientific formulas; calculators handle it directly.
• Use log if you prefer base 10 or if your calculator defaults to it.
• Either works: x=ln(20)/ln(7) equals x=log(20)/log(7).

Using a convenient base (change-of-base idea in practice)

Sometimes the cleanest expression is a base that matches the exponential base:

• If you have 5^{x}=12, writing x=log_{5}(12) is compact.
• On a calculator, you might compute it as x=ln(12)/ln(5) or x=log(12)/log(5).

Quick decision guide

4) Checking solutions and understanding extraneous issues

Exponential equations can produce incorrect candidates if you perform algebraic steps that are not logically reversible (for example, squaring both sides, taking even roots, or introducing absolute values). Taking logs of both sides is generally safe only when both sides are positive, because logarithms are defined for positive inputs.

Always check: domain and substitution

• Positivity for logs: before applying ln or log, ensure the expression you log is > 0.
• Substitute back: plug your solution into the original equation to verify.

Example 9: a candidate that must be rejected (log domain)

Solve 2^{x}-5=0 and 2^{x}-5=-3 (compare).

• For 2^{x}-5=0: 2^{x}=5 ⇒ x=ln 5/ln 2 (valid).
• For 2^{x}-5=-3: 2^{x}=2 ⇒ x=1 (valid). No log issue occurs because the exponential term remains positive.

Now consider ln(2^{x}-5)=1. Here you must require 2^{x}-5>0 before solving.

1. Domain: 2^{x}>5 ⇒ x>ln 5/ln 2.
2. Solve: ln(2^{x}-5)=1 ⇒ 2^{x}-5=e ⇒ 2^{x}=e+5 ⇒ x=ln(e+5)/ln 2.
3. Check domain: e+5>5, so the solution satisfies the requirement.

Example 10: extraneous solution from squaring

Solve 3^{x}= -9.

There is no real solution because 3^{x}>0 for all real x. If someone squares both sides to get 3^{2x}=81 and then finds x=2, that value fails the original equation because 3^{2}=9, not -9. This is a classic extraneous-solution trap caused by squaring.

Example 11: checking after solving with logs

Solve 10(1.3)^{t}=0.5.

1. Isolate: (1.3)^{t}=0.05.
2. Log both sides: t ln(1.3)=ln(0.05).
3. Solve: t=ln(0.05)/ln(1.3) (negative, which makes sense because a growth factor > 1 must be raised to a negative power to produce a small number).
4. Check by substitution conceptually: since t<0, (1.3)^{t} is less than 1, consistent with 0.05.

5) Application problems: time to reach a threshold in growth/decay models

Many real questions ask: “How long until the quantity reaches a certain level?” These are threshold problems. The variable (time) is in the exponent, so you typically isolate the exponential part and use logs.

Application 1: bacterial growth to a target

A culture starts with 500 bacteria and grows by 18% per hour. When will it reach 2000 bacteria?

500(1.18)^t = 2000

1. Divide by 500: (1.18)^t=4.
2. Take logs: t ln(1.18)=ln 4.
3. Solve: t=ln 4/ln(1.18) hours.
4. Interpretation: the answer is the time to hit the threshold; if you need the first whole hour exceeding 2000, round up after evaluating.

Application 2: depreciation to a threshold value

A machine costs $12,000 and loses 9% of its value each year. After how many years is it worth $6,000?

12000(0.91)^t = 6000

1. Divide by 12000: (0.91)^t=0.5.
2. Log both sides: t ln(0.91)=ln(0.5).
3. Solve: t=ln(0.5)/ln(0.91) years (positive because both logs are negative).

Application 3: half-life style threshold

A substance decays according to A(t)=A_0(0.84)^t where t is in days. How long until only 10% remains?

A0(0.84)^t = 0.10 A0

1. Cancel A0: (0.84)^t=0.10.
2. Take logs: t ln(0.84)=ln(0.10).
3. Solve: t=ln(0.10)/ln(0.84) days.

Application 4: continuous growth threshold (using e)

An investment follows V(t)=2500e^{0.06t} where t is in years. When does it reach $4000?

2500e^(0.06t) = 4000

1. Divide by 2500: e^{0.06t}=1.6.
2. Take ln: ln(e^{0.06t})=ln(1.6).
3. Simplify: 0.06t=ln(1.6).
4. Solve: t=ln(1.6)/0.06 years.

Application 5: threshold with a shifted model (isolate carefully)

A medication level is modeled by C(t)=30+120(0.7)^t (mg/L). When does it drop below 50 mg/L?

30 + 120(0.7)^t < 50

1. Isolate the exponential: subtract 30: 120(0.7)^t < 20.
2. Divide by 120: (0.7)^t < 1/6.
3. Take logs: t ln(0.7) < ln(1/6).
4. Important inequality note: since ln(0.7)<0, dividing flips the inequality: t > ln(1/6)/ln(0.7).
5. Interpretation: any time greater than that value yields concentration below 50 mg/L.