Solve: (1/5)log₈1024 - 2log₈(1/2) Logarithmic Expression

Logarithmic Properties with Power Rules

15log810242log812= \frac{1}{5}\log_81024-2\log_8\frac{1}{2}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:08 We'll use the power logarithm formula
00:12 We'll raise the numbers to their appropriate powers
00:23 We'll calculate each power
00:53 We'll use the logarithm subtraction formula
00:57 Subtracting logarithms equals the logarithm of the quotient
01:03 We'll use this formula in our exercise
01:10 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

15log810242log812= \frac{1}{5}\log_81024-2\log_8\frac{1}{2}=

2

Step-by-step solution

To solve this problem, we'll begin by simplifying the given expression using logarithmic rules:

  • Step 1: Simplify the first term 15log81024\frac{1}{5}\log_8 1024:
    Since 1024=2101024 = 2^{10}, we can rewrite this as log8(210)=10log82\log_8 (2^{10}) = 10 \log_8 2. Thus, 15log81024=15(10log82)=2log82\frac{1}{5} \log_8 1024 = \frac{1}{5} (10 \log_8 2) = 2 \log_8 2.
  • Step 2: Simplify the second term 2log8122 \log_8 \frac{1}{2}:
    Note 12=21\frac{1}{2} = 2^{-1}. Therefore, log8(12)=log8(21)=1log82\log_8 \left(\frac{1}{2}\right) = \log_8(2^{-1}) = -1 \log_8 2. Thus, 2log8(12)=2(log82)=2log822 \log_8 \left(\frac{1}{2}\right) = 2(-\log_8 2) = -2 \log_8 2.
  • Step 3: Subtract the expressions:
    Combine the terms using the difference rule:
    2log82(2log82)=2log82+2log82=4log822 \log_8 2 - (-2 \log_8 2) = 2 \log_8 2 + 2 \log_8 2 = 4 \log_8 2.
  • Step 4: Simplify further:
    Since 4=224 = 2^2, we can express this as log8(24)=log8(16)\log_8 (2^4) = \log_8 (16).

Therefore, the simplified form of the expression is log816\log_8 16.

The correct choice is thus log816\log_8 16, matching with choice (1).

3

Final Answer

log816 \log_816

Key Points to Remember

Essential concepts to master this topic
  • Power Rule: Use logarithm property logb(an)=nlogb(a) \log_b(a^n) = n\log_b(a)
  • Technique: Convert 1024 = 2^{10} and 1/2 = 2^{-1} to use common base
  • Check: Final answer log816 \log_8 16 where 16 = 2^4 makes sense ✓

Common Mistakes

Avoid these frequent errors
  • Not recognizing exponential forms
    Don't work with 1024 and 1/2 directly without converting to powers of 2 = complicated calculations! Students miss that 1024 = 2^{10} and 1/2 = 2^{-1}, making the problem much harder. Always look for common bases like powers of 2 first.

Practice Quiz

Test your knowledge with interactive questions

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

FAQ

Everything you need to know about this question

How do I know that 1024 equals 2 to the 10th power?

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Start by repeatedly dividing by 2: 1024 ÷ 2 = 512, 512 ÷ 2 = 256, and so on. Count how many times you can divide by 2 until you reach 1. That's your exponent!

Why does the negative sign appear when I have 1/2?

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Because 12=21 \frac{1}{2} = 2^{-1} ! When you take the logarithm of negative exponents, the negative comes out front: log8(21)=1log8(2) \log_8(2^{-1}) = -1 \cdot \log_8(2) .

What's the difference between subtraction and addition of logarithms?

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Subtraction becomes division: logb(a)logb(c)=logb(ac) \log_b(a) - \log_b(c) = \log_b(\frac{a}{c})
Addition becomes multiplication: logb(a)+logb(c)=logb(ac) \log_b(a) + \log_b(c) = \log_b(ac)

How do I combine terms with coefficients like 2log₈(2)?

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The coefficient becomes an exponent inside the logarithm: 2log8(2)=log8(22)=log8(4) 2\log_8(2) = \log_8(2^2) = \log_8(4) . This makes combining terms much easier!

Can I just calculate the decimal values instead?

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Not recommended! Working with exact logarithmic forms like log816 \log_8 16 is more accurate than decimals. Plus, many problems expect the exact logarithmic answer.

What if I get confused with all the logarithm rules?

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Focus on three main rules:

  • Power rule: logb(an)=nlogb(a) \log_b(a^n) = n\log_b(a)
  • Product: logb(xy)=logb(x)+logb(y) \log_b(xy) = \log_b(x) + \log_b(y)
  • Quotient: logb(xy)=logb(x)logb(y) \log_b(\frac{x}{y}) = \log_b(x) - \log_b(y)

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