Solve the Logarithmic Equation: log₄(x²)·log₇16 = 2log₇8

Logarithmic Equations with Change of Base

log4x2log716=2log78 \log_4x^2\cdot\log_716=2\log_78

?=x

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:03 Find the domain of definition
00:16 We'll use the formula for logarithmic multiplication, we'll switch between the bases
00:31 We'll use this formula in our exercise
00:41 Let's calculate the logarithm
00:56 We'll substitute in our exercise and continue solving
01:06 Let's simplify what we can
01:16 We'll compare the numbers, since the bases are equal
01:21 When extracting a root there are always 2 solutions, positive and negative
01:25 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

log4x2log716=2log78 \log_4x^2\cdot\log_716=2\log_78

?=x

2

Step-by-step solution

To solve this logarithmic equation, we will break down and simplify the given expression step by step:

Step 1: Simplify each logarithm using the change of base formula.

First, consider log4x2 \log_4 x^2 :
Using the power rule, log4x2=2log4x\log_4 x^2 = 2 \log_4 x.
Now apply the change of base formula:
log4x=logxlog4\log_4 x = \frac{\log x}{\log 4}, thus log4x2=2logxlog4\log_4 x^2 = 2 \cdot \frac{\log x}{\log 4}.

Step 2: Simplify log716\log_7 16 and log78\log_7 8 using the change of base formula.
log716=log16log7=log(24)log7=4log2log7\log_7 16 = \frac{\log 16}{\log 7} = \frac{\log (2^4)}{\log 7} = \frac{4 \log 2}{\log 7}.
Similarly, log78=log8log7=log(23)log7=3log2log7\log_7 8 = \frac{\log 8}{\log 7} = \frac{\log (2^3)}{\log 7} = \frac{3 \log 2}{\log 7}.

Step 3: Substitute these values back into the equation.
2logxlog44log2log7=23log2log7 \frac{2 \log x}{\log 4} \cdot \frac{4 \log 2}{\log 7} = 2 \cdot \frac{3 \log 2}{\log 7}

Step 4: Simplify the equation by canceling out common terms and solving for logx\log x.
After cancelling log2log7\frac{\log 2}{\log 7} from both sides, we have:
8logxlog4=6\frac{8 \log x}{\log 4} = 6.

Step 5: Calculate log4=2log2\log 4 = 2 \log 2, so substitute:
8logx2log2=6    4logx=6log2\frac{8 \log x}{2 \log 2} = 6 \implies 4 \log x = 6 \log 2, thus logx=32log2\log x = \frac{3}{2} \log 2.

Step 6: Solve for xx using exponentiation.
Since logx=32log2\log x = \frac{3}{2} \log 2, exponentiation gives x=232=8x = 2^{\frac{3}{2}} = \sqrt{8}. However, since logarithms are defined for positive numbers, we must consider ±\pm for solutions within the constraints. Thus, x=±8x = \pm \sqrt{8}.

Therefore, the solution to the problem is x=±8 x = \pm\sqrt{8} , corresponding to choice 44.

3

Final Answer

±8 \pm\sqrt{8}

Key Points to Remember

Essential concepts to master this topic
  • Power Rule: Use log(a^n) = n·log(a) to simplify logarithms
  • Change of Base: Convert log₄(x²) = 2log(x)/log(4) for easier calculation
  • Check: Verify by substituting x = ±√8 back into original equation ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting to consider both positive and negative solutions
    Don't just take x = √8 as the only answer = missing half the solutions! Since x² appears in log₄(x²), both positive and negative values of x give the same result. Always consider x = ±√8 when dealing with squared variables.

Practice Quiz

Test your knowledge with interactive questions

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

FAQ

Everything you need to know about this question

Why do we need the change of base formula here?

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The change of base formula converts different logarithm bases to common logarithms, making it easier to simplify and solve. Without it, comparing log4 \log_4 and log7 \log_7 would be very difficult!

How do I know when to use ± in my answer?

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Look for squared variables like x2 x^2 in the original equation. Since both positive and negative numbers give the same result when squared, you need both solutions: x = ±√8.

What's the difference between log₄(x²) and 2log₄(x)?

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They're equal! The power rule states that log4(x2)=2log4(x) \log_4(x^2) = 2\log_4(x) . This is a key property that helps simplify logarithmic expressions.

Why can't I just solve this without changing the base?

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Different bases make comparison impossible. The change of base formula converts everything to common logarithms, allowing you to cancel terms and solve algebraically.

How do I verify my answer is correct?

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Substitute x=±8 x = \pm\sqrt{8} back into the original equation. Both sides should equal the same value. Remember: (8)2=(8)2=8 (\sqrt{8})^2 = (-\sqrt{8})^2 = 8 !

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