Exponential and Logarithmic Equations and Inequalities
Introduced| English | Meaning |
|---|---|
| exponential equation/ˌekspəˈnenʃl ɪˈkweɪʒn/ | An equation in which an unknown occurs in an exponent. |
| domain/dəˈmeɪn/ | The set of input values for which a function is defined. |
| logarithmic equation/ˌlɒɡəˈrɪθmɪk ɪˈkweɪʒn/ | logarithmic equation |
| extraneous/ekˈstreɪnɪəs/ | extraneous |
| inequality/ɪniːˈkwɒlɪti/ | A mathematical statement comparing expressions using relations such as less than or greater than. |
| inverse/ɪnˈvɜːs/ | A function that reverses another function's mapping on a suitable one-to-one domain. |
Recover a time from a target
- A model $N(t)=100\cdot2^t$ starts with 100 and doubles each hour. Reaching 500 requires $2^t=5$: the unknown is an exponent, not a factor that can simply be divided out.
- An algebraic answer still needs interpretation. If the model starts at time zero, only $t\ge0$ is allowed. Keep that restriction when deciding whether a calculated time answers the question.
Match bases when that simplifies the equation
- An exponential equation 指数方程 has a variable in an exponent. Since $8=2^3$, the equation $2^x=8$ gives $x=3$. For a valid base $b>0$, $b\ne1$, equal powers have equal exponents.
- For $9^{x-1}=27$, use base 3: $3^{2(x-1)}=3^3$. Thus $2(x-1)=3$ and $x=2.5$. Check the entire exponent, including its shift.
Solve $2^x = 8$ by matching bases.
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$8 = 2^3$, so $2^x = 2^3$ gives $x = 3$ — equal bases mean equal exponents.
Isolate the exponential term first
- To solve $3\cdot2^{x-1}+4=28$, subtract 4 and divide by 3, giving $2^{x-1}=8$. Then $x-1=3$, so $x=4$.
- Check whether the isolated right side can be an exponential output. An equation such as $2^x=-1$ has no real solution, because $2^x>0$ for every real $x$. Taking a real logarithm of $-1$ is not a valid next step.
Solve $3\cdot 2^{x-1}+4=28$.
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Subtract 4 and divide by 3 to get 2^(x −1)=8. Then x −1=3, so x=4.
Use logarithms for an unfamiliar power
- For $2^x=5$, applying the matching logarithm gives $x=\log_2 5=\dfrac{\ln5}{\ln2}\approx2.322$. Keep the logarithmic expression as the exact answer and round only the final approximation.
- Substitution checks $2^{2.322}\approx5$. Graphically, this is the input where $y=2^x$ meets $y=5$. A plotted crossing estimates the answer; the logarithm specifies it exactly.
To solve $2^x=5$ without a convenient common-base rewrite, you should…
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Apply a matching logarithm: x=log₂5=ln(5)/ln(2)≈2.322. Retain the exact logarithmic expression until the final approximation.
Isolate a logarithm and keep its domain
- A logarithmic equation 对数方程 can be rewritten in exponential form: $\log_2 x=4$ gives $x=2^4=16$. Its domain 定义域 requires $x>0$, which this answer satisfies.
- For $3\log_2(x-2)+5=11$, first require $x>2$. Isolating gives $\log_2(x-2)=2$, so $x-2=4$ and $x=6$. Substitute back into the original equation, not just the last rearranged line.
Solve the logarithmic equation $\log_2 x = 4$.
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Rewrite in exponential form: $x = 2^4 = 16$.
Reject candidates outside the original domain
- For common logarithms, $\log x+\log(x-3)=1$ requires $x>3$. Condensing gives $x(x-3)=10$, hence $(x-5)(x+2)=0$. The candidates are 5 and $-2$.
- Only 5 is valid: $\log5+\log2=\log10=1$. The candidate $-2$ is extraneous 增根 because its original log arguments are negative. A zero argument would also be excluded.
A candidate solution that makes an original log argument nonpositive must be rejected as ____.
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An extraneous candidate may satisfy a transformed equation while failing the original domain. Check every original log argument, including exclusion of zero.
Solving $\log x+\log(x-3)=1$ gives candidates 5 and −2. Which are valid in the original equation?
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Both original arguments must be positive, so x>3. Only 5 survives, and log 5+log 2=1.
Equate log arguments only after checking them
- With a valid common base, logarithms are one-to-one, even when they decrease. Thus $\log_b U=\log_b V$ implies $U=V$, provided $U,V>0$.
- For $\log_2(x+3)=\log_2(2x-1)$, the original domain is $x>1/2$. Equating arguments gives $x+3=2x-1$, so $x=4$. Both arguments then equal 7, which confirms the solution.
Select all true statements about solving these equations.
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Some candidates are extraneous and must be rejected. The other three are sound strategies.
Solve $\log_2(x+3)=\log_2(2x-1)$.
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The domain is x>1/2. Equal logs with the same valid base give x+3=2x −1, so x=4, with both arguments 7.
Use the base to decide an inequality's direction
- An inequality 不等式 needs monotonicity as well as a boundary. Since $2^x$ increases, $2^x>8$ gives $x>3$. The boundary is excluded because the comparison is strict.
- Since $(1/2)^x$ decreases, $(1/2)^x\ge1/8$ gives $x\le3$. The direction reverses, and equality is included. Testing $x=0$ gives $1\ge1/8$, confirming the chosen side.
For $-2\cdot2^x<-16$, dividing by $-2$ reverses the comparison to $2^x>8$. Since base 2 is increasing, the solution is $x>3$. A negative coefficient can reverse the sign before the exponential comparison.
Solve $(1/2)^x\ge 1/8$.
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The equality boundary is 3, and the decreasing function reverses the comparison. Equality is included.
Solve $-2\cdot2^x<-16$.
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Dividing by −2 reverses the comparison to 2ˣ>8. Base 2 is increasing, so x>3. The strict comparison excludes 3.
Intersect a log inequality with its domain
- For $\log_2(x-1)\ge2$, start with $x>1$. The increasing base gives $x-1\ge4$, so the solution is $x\ge5$.
- For $\log_{1/2}(x-1)\ge-2$, retain $x>1$ and reverse the comparison: $x-1\le4$. The full answer is $1<x\le5$. Changing the original comparison to $>$ gives $1<x<5$ instead.
Read the boundary, domain and solution set
Start with 2ˣ=5. Compare increasing and decreasing relations, strict and included boundaries, and cases with no crossing.
Solve $\log_{1/2}x>-3$.
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Require x>0. The decreasing base reverses the inequality, so x<(1/2)^(−3)=8. The strict comparison excludes 8.
Solve $\log_{1/2}(x-1)\ge-2$.
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The domain requires x>1. The decreasing base gives x −1≤4. Intersect the two conditions to obtain 1<x≤5.
Check mathematical and contextual limitations
- The equation $100\cdot2^t=50$ gives $t=-1$. This is mathematically valid, but it is outside the model's stipulated time interval $t\ge0$. The target is never reached after the stated start.
- Also check whether a boundary exists at all. The equation $2^x=-1$ has no real solution, while the inequality $2^x>-1$ holds for every real $x$. An equation's lack of a crossing does not mean every related inequality has no solution.
The equation $2^x=-1$ has a real solution that can be found by taking a real logarithm of −1.
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2ˣ is strictly positive for every real x. There is no real solution and no valid real log of −1.
A model $N(t)=100\cdot 2^t$ is restricted to $t\ge 0$. When does it reach 50?
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The algebraic time is −1, but this is outside t≥0. Since the model increases from 100, it cannot reach 50 after the start.
Rewrite an exponential with another base
- For valid bases $b$ and $c$, $b^x=c^{x\log_c b}$. In particular, $8^x=2^{3x}$ because $\log_2 8=3$.
- The natural-base version is $b^x=e^{x\ln b}$. Thus $3\cdot8^x=3e^{x\ln8}$, exposing the same initial value 3 and the continuous exponent coefficient $\ln8$. Changing the base preserves the function.
Write $8^x=2^{kx}$ for every real x. What is k?
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8=2³, so 8ˣ=(2³)ˣ=2^(3x). Equivalently,k=log₂8=3.
Reverse a transformed exponential
- For $f(x)=ab^{x-h}+k$, with $a\ne0$ and a valid base, solve $y-k=ab^{x-h}$ by dividing by $a$ and taking $\log_b$. The inverse 反函数 is $f^{-1}(y)=h+\log_b\!\left(\frac{y-k}{a}\right)$.
- Its domain requires $(y-k)/a>0$: $y>k$ when $a>0$, or $y<k$ when $a<0$. For $f(x)=3\cdot2^{x-1}+4$, the inverse is $1+\log_2((y-4)/3)$ with domain $y>4$. Its range is all real numbers.
For $f(x)=3\cdot 2^{x-1}+4$, which inverse rule and domain are correct?
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Subtract 4, divide by 3, take log₂, then add 1. The log argument (y −4)/3 must be positive, giving y>4.
Reverse a transformed logarithm
- For $f(x)=a\log_b(x-h)+k$, with $a\ne0$, start with domain $x>h$. Isolate the logarithm to obtain $(y-k)/a=\log_b(x-h)$.
- Exponentiating gives $f^{-1}(y)=h+b^{(y-k)/a}$, with domain all real numbers and range greater than $h$. For $f(x)=3\log_2(x-2)+5$, the inverse is $2+2^{(y-5)/3}$; it sends 11 back to 6, and always outputs a value greater than 2.
Isolate before applying an inverse, keep exact forms until the final approximation, and check the original domain and context. For inequalities, use the function's direction, retain excluded domain edges and distinguish open from closed solution boundaries.
For $f(x)=3\log_2(x-2)+5$, which inverse rule and domain are correct?
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Subtract 5, divide by 3, exponentiate base 2 and add 2. The inverse accepts all real inputs and outputs values greater than 2.
Explain a solution and two inequalities
- Solve $3^x=81$ by matching bases, then check the result.
- Compare the solution sets of $3^x>81$ and $(1/3)^x>1/81$. Explain the direction and whether the boundary belongs to either set.
Solve $3^x=81$ and check it. Then solve $3^x>81$ and $(1/3)^x>1/81$, explaining which side of the boundary is valid and whether equality is included.
The exponents match at 4.
Solve $3^x=81$. What is x?
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3 to the fourth power is 81. Other options — 3 (the lesson solution of 2 to the power x equals 8); 16 (the lesson solution of a log equation).