Solve: (1/2)log₃9 - log₃1.5 Logarithmic Expression

Logarithmic Properties with Power Rules

12log39log31.5= \frac{1}{2}\log_39-\log_31.5=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:13 We will use the formula for the log of a power
00:26 We'll convert from power of half to root and calculate
00:38 Now we'll use the formula for subtracting logarithms
00:43 Subtracting logs equals the log of the quotient of numbers
00:47 We'll use this formula in our exercise
01:05 We'll convert from number to fraction and calculate
01:12 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

12log39log31.5= \frac{1}{2}\log_39-\log_31.5=

2

Step-by-step solution

To solve the problem 12log39log31.5 \frac{1}{2}\log_39-\log_31.5 , we need to apply the rules of logarithms:

  • **Step 1: Simplify with the power rule**
    Using the power rule 12log39=log391/2 \frac{1}{2}\log_39 = \log_39^{1/2} . Since 9=329 = 3^2, we have 91/2=321/2=319^{1/2} = 3^{2 \cdot 1/2} = 3^1. Thus, 12log39=log33=1\frac{1}{2}\log_39 = \log_3 3 = 1.
  • **Step 2: Apply the subtraction rule**
    Now, the expression becomes 1log31.51 - \log_3 1.5. Using the subtraction rule: 1log31.5=log33log31.5=log3(31.5)1 - \log_3 1.5 = \log_3 3 - \log_3 1.5 = \log_3 \left(\frac{3}{1.5}\right).
  • **Step 3: Simplify the fraction**
    Calculate 31.5\frac{3}{1.5}: it simplifies to 2 because 3÷1.5=23 \div 1.5 = 2.

Thus, the simplified expression is log32\log_3 2.

Using the provided answer choices, the correct answer matches choice log32 \log_3 2 , which corresponds to choice 2.

Therefore, the solution to the problem is log32 \log_3 2 .

3

Final Answer

log32 \log_32

Key Points to Remember

Essential concepts to master this topic
  • Power Rule: Convert coefficients to exponents: 12log39=log391/2 \frac{1}{2}\log_39 = \log_39^{1/2}
  • Technique: Simplify bases first: 9=32 9 = 3^2 so 91/2=3 9^{1/2} = 3
  • Check: Verify final answer: log32 \log_3 2 means 3log32=2 3^{\log_3 2} = 2

Common Mistakes

Avoid these frequent errors
  • Not recognizing that 9 = 3² before applying the power rule
    Don't leave 12log39 \frac{1}{2}\log_39 as log39=log33=1 \log_3\sqrt{9} = \log_33 = 1 ! Students often miss this crucial simplification step and get stuck with complex expressions. Always identify perfect powers in the base first: since 9=32 9 = 3^2 , then 91/2=3 9^{1/2} = 3 .

Practice Quiz

Test your knowledge with interactive questions

\( \log_{10}3+\log_{10}4= \)

FAQ

Everything you need to know about this question

Why does 12log39 \frac{1}{2}\log_39 equal log391/2 \log_39^{1/2} ?

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This is the power rule of logarithms! When you have a coefficient in front of a log, you can move it as an exponent inside: alogbx=logbxa a\log_b x = \log_b x^a .

How do I know that 91/2=3 9^{1/2} = 3 ?

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Since 9=32 9 = 3^2 , we have 91/2=(32)1/2=321/2=31=3 9^{1/2} = (3^2)^{1/2} = 3^{2 \cdot 1/2} = 3^1 = 3 . The square root of 9 is 3!

Why does log33log31.5=log3(3/1.5) \log_3 3 - \log_3 1.5 = \log_3(3/1.5) ?

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This uses the quotient rule: logaxlogay=loga(x/y) \log_a x - \log_a y = \log_a(x/y) . Since log33=1 \log_3 3 = 1 , we can write 1 as log33 \log_3 3 and apply the rule.

How does 3÷1.5=2 3 ÷ 1.5 = 2 ?

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Convert the decimal: 1.5=32 1.5 = \frac{3}{2} . So 3÷1.5=3÷32=3×23=2 3 ÷ 1.5 = 3 ÷ \frac{3}{2} = 3 \times \frac{2}{3} = 2 . Always simplify fractions step by step!

Can I check if log32 \log_3 2 is correct?

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Yes! log32 \log_3 2 means "what power gives me 2 when I raise 3 to it?" Since 30.63...2 3^{0.63...} ≈ 2 , this confirms our answer is reasonable.

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