Manipulating Logarithmic Expressions
Introduced| English | Meaning |
|---|---|
| equivalent/ɪˈkwɪvələnt/ | Having the same value, meaning or effect in the relationship being considered. |
| domain/dəˈmeɪn/ | The set of input values for which a function is defined. |
| product rule/ˈprɒdʌkt ruːl/ | The differentiation rule for two multiplied functions: the first times the derivative of the second plus the second times the derivative of the first. |
| quotient rule/ˈkwəʊʃənt ruːl/ | The differentiation rule for a ratio: the denominator times the numerator's derivative minus the numerator times the denominator's derivative, divided by the denominator squared. |
| power rule/ˈpaʊə ruːl/ | The differentiation rule stating that the derivative of x raised to n is n multiplied by x raised to (n - 1). |
| absolute value/ˈæbsəluːt ˈvæljuː/ | absolute value |
| change-of-base/tʃeɪndʒ ɒv beɪs/ | Rewriting a logarithm using logarithms in another base. |
Combine multiplication steps
- Two growth multipliers, 8 and 4, combine to give a factor of 32. In base 2, they represent 3 and 2 doubling steps, so the combined number of steps is $3+2=5$.
- This is the idea behind $\log_2(8\cdot4)=\log_2 8+\log_2 4$. Multiplication of positive quantities becomes addition of their logarithms.
State the restrictions before rewriting
- For real logarithms, use $b>0$, $b\ne1$ and positive arguments. The standard product, quotient and power identities below assume $x>0$ and $y>0$; a real power $n$ is allowed when its base $x$ is positive.
- Two forms are equivalent 等价 on the stated domain 定义域 when they give the same values there. A rewrite must retain the original restrictions, even if the new formula is defined at extra inputs.
Use the product rule
- The product rule 乘法法则 is $\log_b(xy)=\log_b x+\log_b y$, for $x,y>0$. In the opening example, $\log_2(8\cdot4)=3+2=5$.
- To see why, write $x=b^u$ and $y=b^v$. Then $xy=b^{u+v}$, so its logarithm is $u+v$. It is the sum of the two exponents, not their product.
For $x,y>0$ and a valid base b, the product rule says $\log_b(xy)=$
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A log of a product is a sum of logs — the mirror of "multiplying powers adds exponents".
Use the quotient rule
- The quotient rule 除法法则 is $\log_b(x/y)=\log_b x-\log_b y$, for $x,y>0$. For example, $\log_2(8/4)=3-2=1$, which agrees with $\log_2 2=1$.
- Dividing the arguments subtracts their logarithms. It does not divide the logarithms. If $x$ and $y$ are both negative, the quotient can be positive while the two separate real logarithms are undefined.
For $x,y>0$, the quotient rule says $\log_b(x/y)=\dfrac{\log_b x}{\log_b y}$.
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False. The quotient rule subtracts: log_b(x/y)=log_b(x)−log_b(y), with x,y>0 and a valid common base.
Evaluate $\log_2(8/4)$.
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The quotient rule gives 3−2=1. Equivalently, 8/4=2 and log₂2=1.
Use the power rule on positive inputs
- The power rule 幂法则 is $\log_b(x^n)=n\log_b x$ for $x>0$. Thus $\log_2(8^3)=3\log_2 8=9$.
- Fractional and negative powers work too: $\log_2\sqrt8=\tfrac12\log_2 8=1.5$, while $\log_2(8^{-1})=-\log_2 8=-3$. The exponent becomes a multiplier, not a power of the logarithm.
For $x>0$, real n and a valid base b, the power rule says $\log_b(x^n)=$
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On x>0, the exponent becomes a multiplier: log_b(x^n)=n log_b(x). Retain this positive-input restriction when using that form.
Evaluate $\log_2\sqrt8$.
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√8=8^(1/2), so its logarithm is (1/2)log₂8=1.5.
Keep the domain when a square hides a sign
- The expression $\log_b(x^2)$ is defined for $x\ne0$, but $2\log_b x$ requires $x>0$. At $x=-2$, the first can be evaluated; the second cannot. They are equivalent only on their common positive domain.
- The absolute value 绝对值 gives the full real-domain identity $\log_b(x^2)=2\log_b|x|$ for $x\ne0$. At $x=-2$ and $b=2$, both sides of this corrected identity equal 2.
Compare logarithm laws and their domains
Start with log₂(8×4)=3+2=5. Compare valid identities, then check negative inputs and the square rule.
At $x=-2$, which statement is correct?
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The square gives positive argument 4. The separate log₂x still has negative argument. The full-domain identity uses 2log₂|x|.
Expand and condense with the same restrictions
- For $x,y,z>0$, condense $2\log x+\log y-\log z$ into $\log(x^2y/z).$ Retain all three positivity restrictions: the condensed expression alone can accept inputs that the original does not.
- Expanding can also separate a constant, powers, products and quotients. For positive $x,y,z$, the following expression becomes a sum and differences. Keep the square-root factor as a power of $1/2$.
After condensing $2\log x+\log y-\log z$ into $\log(x^2y/z)$, which restrictions must be retained?
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The original separate logarithms need all three inputs positive. The condensed formula alone has a larger domain.
Do not split sums inside a logarithm
- The product rule does not say $\log_b(x+y)=\log_b x+\log_b y$. For example, $\log_2(8+8)=4$, whereas $\log_2 8+\log_2 8=6$.
- Subtraction inside an argument does not split either: $\log_2(8-4)=2$, while $\log_2 8-\log_2 4=1$. Identify the operation inside the parentheses before choosing a law.
For $x,y>0$ and real n, select all correct common-log laws.
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There is no rule for $\log(x+y)$ — the laws apply to products, quotients, and powers, not sums. The other three are correct.
Evaluate $\log_2(8+8)$.
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The argument is 16, so the answer is 4. Adding the two separate logarithms would give 6 and is not a valid expansion of this sum.
Interpret the laws as graph transformations
- For $k>0$, $\log_b(kx)=\log_b k+\log_b x$ on $x>0$. For example, $\log_2(4x)=2+\log_2 x$: a horizontal compression by factor $1/4$ gives the same graph as a vertical shift up by 2.
- On $x>0$ with $n\ne0$, $\log_b(x^n)=n\log_b x$ is a vertical dilation by factor $|n|$, with reflection across the x-axis when $n<0$. For $n=2$, extending the left expression to negative $x$ adds a branch that is not part of $2\log_b x$.
On $x>0$, which vertical transformation gives the graph of $\log_2(4x)$ from $\log_2 x$?
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log₂(4x)=log₂4+log₂x=2+log₂x. This also represents a horizontal compression by factor 1/4.
The graphs of $\log_2(x^2)$ and $2\log_2 x$ are identical on all nonzero real inputs.
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They agree on x>0. The left expression has an additional negative-input branch; the right expression is undefined there.
Change to a convenient valid base
- The change-of-base 换底 formula is $\log_b x=\dfrac{\log_a x}{\log_a b}$, with $x>0$ and both bases positive and different from 1. The denominator is nonzero because $b\ne1$.
- With LOG or LN keys, use $\log_2 7=\dfrac{\ln7}{\ln2}\approx2.807$. The order matters: argument on top, original base below. Check that $2^{2.807}\approx7$, keeping extra digits until the final answer.
The ____ formula lets you compute $\log_2 7$ as $\dfrac{\log 7}{\log 2}$ on a calculator.
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Use log_b(x)=log_a(x)/log_a(b), where x>0 and a,b are positive and different from 1. The argument is in the numerator; the original base is in the denominator.
Connect different logarithmic bases
- The formula $\log_b x=\dfrac{1}{\ln b}\ln x$ shows a vertical dilation of $\ln x$ on $x>0$. The multiplier is positive for $b>1$ and negative for $0<b<1$; a negative multiplier includes a reflection.
- For example, $\log_{1/2}x=\dfrac{\log_2 x}{\log_2(1/2)}=-\log_2 x$. In particular, $\log_{1/2}8=-3$. Changing base preserves the positive domain.
Use positive arguments and a valid common base for the standard laws. Products give sums, quotients give differences and powers give multipliers. Preserve the original domain when expanding or condensing, and use change of base with the numerator and denominator in the correct order.
Use change of base to evaluate $\log_{1/2}8$.
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log₂8/log₂(1/2)=3/(−1)=−3. A base below 1 gives a negative vertical multiplier relative to log₂x.
Explain both the value and the domain
- Use the product rule to evaluate $\log_2(8\cdot4)$, and check the result by evaluating the product first.
- If both factors were negative instead, could you always split the logarithm of their product into two real logarithms? Give a numerical example and explain the restrictions.
Use the product rule to evaluate $\log_2(8\cdot4)$ and check by multiplying first. Then explain whether $\log_2((-8)(-4))$ can be split into $\log_2(-8)+\log_2(-4)$.
3 plus 2 is 5.
What is $\log_2 8+\log_2 4$?
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The product rule adds the logs.