Solve the Logarithmic Equation: log₃(x²)log₅(27) - log₅(8) = ln(e)

Logarithmic Equations with Mixed Base Conversion

log3x2log527log58=lne \log_3x^2\log_527-\log_58=\ln e

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:09 We'll use the logarithm product formula, switching between bases
00:24 We'll use this formula in our exercise
00:44 Let's calculate the logarithm
00:59 We'll substitute this solution in our exercise and continue solving
01:29 We'll use the power logarithm formula
01:34 We'll use this formula in our exercise
01:54 We'll substitute in our exercise
02:09 We'll use the logarithm quotient formula, we'll get the logarithm of the quotient
02:24 We'll use this formula in our exercise
02:39 We'll convert to logarithm and calculate
02:49 We'll solve according to the logarithm definition
03:09 Let's isolate X
03:34 This is the possible solution, now let's check the domain of definition
03:45 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

log3x2log527log58=lne \log_3x^2\log_527-\log_58=\ln e

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Convert the logarithms into another base using the change of base rule.
  • Step 2: Simplify lne\ln e since lne=1\ln e = 1.
  • Step 3: Simplify the expression using known values.
  • Step 4: Solve the equation for x x .

Now, let's work through each step:

Step 1: Given the equation log3x2log527log58=lne \log_3 x^2 \log_5 27 - \log_5 8 = \ln e , we know that lne=1\ln e = 1. We will first simplify the right side to get:
log3x2log527log58=1 \log_3 x^2 \log_5 27 - \log_5 8 = 1

Step 2: Use the change of base formula.

Using logba=lnalnb\log_b a = \frac{\ln a}{\ln b}, rewrite log527 \log_5 27 and log58 \log_5 8 :

log527=ln27ln5andlog58=ln8ln5 \log_5 27 = \frac{\ln 27}{\ln 5} \quad \text{and} \quad \log_5 8 = \frac{\ln 8}{\ln 5}

Plug in the values:

log3x2ln27ln5ln8ln5=1 \log_3 x^2 \frac{\ln 27}{\ln 5} - \frac{\ln 8}{\ln 5} = 1

Step 3: Multiply through by ln5 \ln 5 to eliminate the denominators:
log3x2ln27ln8=ln5 \log_3 x^2 \ln 27 - \ln 8 = \ln 5

Now knowing ln27=3ln3\ln 27 = 3\ln 3, solve the equation:

log3x2=ln5+ln83ln3 \log_3 x^2 = \frac{\ln 5 + \ln 8}{3 \ln 3}

Apply the logarithm base rule:

x2=3(ln5+ln83ln3) x^2 = 3^{\left(\frac{\ln 5 + \ln 8}{3\ln 3}\right)}

Step 4: Simplify and solve for x x . Recognize this exponent could become ln403ln3\frac{\ln 40}{3\ln 3}:

x2=3ln403ln3=401/3 x^2 = 3^{\frac{\ln 40}{3\ln 3}} = 40^{1/3}

Finally, solve for x x :

x=±406 x = \pm \sqrt[6]{40}

Therefore, the solution to the problem is x=±406 x = \pm\sqrt[6]{40} .

3

Final Answer

±406 \pm\sqrt[6]{40}

Key Points to Remember

Essential concepts to master this topic
  • Base Conversion: Use change of base formula to work with natural logarithms
  • Technique: Convert log527=ln27ln5=3ln3ln5 \log_5 27 = \frac{\ln 27}{\ln 5} = \frac{3\ln 3}{\ln 5}
  • Check: Substitute x=406 x = \sqrt[6]{40} back into original equation ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting to apply logarithm properties before solving
    Don't leave log3x2 \log_3 x^2 unchanged = impossible to solve directly! Students often skip simplifying logarithmic expressions using properties like logban=nlogba \log_b a^n = n\log_b a . Always convert to the same base and apply logarithm properties first.

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 to convert all logarithms to the same base?

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Different bases make it impossible to combine logarithmic terms directly. Converting to natural logarithms using logba=lnalnb \log_b a = \frac{\ln a}{\ln b} lets you work with familiar operations.

How do I know lne=1 \ln e = 1 ?

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By definition, lne \ln e means "what power do I raise e to get e?" The answer is 1, since e1=e e^1 = e . This is a fundamental property you should memorize!

Why does x2=401/3 x^2 = 40^{1/3} give us x=±406 x = \pm\sqrt[6]{40} ?

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Taking the square root of both sides: x=±401/3 x = \pm\sqrt{40^{1/3}} . Using the property a1/3=a1/6 \sqrt{a^{1/3}} = a^{1/6} , we get x=±406 x = \pm\sqrt[6]{40} .

Can I solve this without the change of base formula?

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Not easily! Mixed bases like log3 \log_3 and log5 \log_5 need conversion to combine terms. The change of base formula is your most reliable tool for these problems.

What if I get confused with all the logarithm properties?

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  • Power rule: logban=nlogba \log_b a^n = n\log_b a
  • Change of base: logba=lnalnb \log_b a = \frac{\ln a}{\ln b}
  • Natural log of e: lne=1 \ln e = 1

Focus on these three key properties for most logarithmic equations!

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