Solve: log₂(3x) × log₅(8) = log₅(a) + log₅(2a) for Variable X

Logarithmic Equations with Change of Base

log⁡23x×log⁡58=log⁡5a+log⁡52a \log_23x\times\log_58=\log_5a+\log_52a

Given a>0 , express X by a

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:15 Let's express X using A.
00:22 We’ll use the logarithm product formula. And remember to switch between bases when needed.
00:37 Now, let's go ahead and calculate the logarithm.
00:47 Next, substitute this solution in. And keep solving the problem!
01:02 Remember to use the formula for a logarithm of a power.
01:12 Now, compare the numbers carefully.
01:22 Let's isolate X, making it stand alone on one side of the equation.
01:37 Great job! That’s how we find the solution to this question.

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

log⁡23x×log⁡58=log⁡5a+log⁡52a \log_23x\times\log_58=\log_5a+\log_52a

Given a>0 , express X by a

2

Step-by-step solution

Let's solve the problem step-by-step:

We start with the equation:

log⁡23x×log⁡58=log⁡5a+log⁡52a \log_2 3x \times \log_5 8 = \log_5 a + \log_5 2a

We simplify the right side using the product rule for logarithms:

log⁡5a+log⁡52a=log⁡5(a⋅2a)=log⁡5(2a2) \log_5 a + \log_5 2a = \log_5 (a \cdot 2a) = \log_5 (2a^2)

Next, we simplify log⁡58\log_5 8 on the left side:

log⁡58=log⁡5(23)=3log⁡52 \log_5 8 = \log_5 (2^3) = 3 \log_5 2

Thus, we substitute into the original equation:

log⁡23x×3log⁡52=log⁡5(2a2) \log_2 3x \times 3 \log_5 2 = \log_5 (2a^2)

Now, divide both sides by 3log⁡523 \log_5 2:

log⁡23x=log⁡5(2a2)3log⁡52 \log_2 3x = \frac{\log_5 (2a^2)}{3 \log_5 2}

Using the change of base formula, express log⁡5(2a2)\log_5 (2a^2) and log⁡52\log_5 2 with base 2:

log⁡5(2a2)=log⁡2(2a2)log⁡25 \log_5 (2a^2) = \frac{\log_2 (2a^2)}{\log_2 5} log⁡52=log⁡22log⁡25=1log⁡25 \log_5 2 = \frac{\log_2 2}{\log_2 5} = \frac{1}{\log_2 5}

Substitute these into the equation:

log⁡23x=log⁡2(2a2)3 \log_2 3x = \frac{\log_2 (2a^2)}{3}

This implies:

log⁡23x=13log⁡2(2a2) \log_2 3x = \frac{1}{3} \log_2 (2a^2)

Raising 2 to both sides of the equation to remove the logarithms:

3x=(2a2)13 3x = (2a^2)^{\frac{1}{3}}

Therefore, solving for x x :

x=13(2a2)13=13⋅2a23 x = \frac{1}{3} (2a^2)^{\frac{1}{3}} = \frac{1}{3} \cdot \sqrt[3]{2a^2}

Thus, we conclude:

x=2a2273 x = \sqrt[3]{\frac{2a^2}{27}}

Therefore, the value of x x in terms of a a is 2a2273 \sqrt[3]{\frac{2a^2}{27}} .

3

Final Answer

2a2273 \sqrt[3]{\frac{2a^2}{27}}

Key Points to Remember

Essential concepts to master this topic
  • Product Rule: Combine logarithms with same base using addition
  • Change of Base: Convert log⁡52=1log⁡25 \log_5 2 = \frac{1}{\log_2 5} to simplify fractions
  • Check: Substitute x=2a2273 x = \sqrt[3]{\frac{2a^2}{27}} back into original equation ✓

Common Mistakes

Avoid these frequent errors
  • Mixing different logarithm bases in operations
    Don't add log⁡23x+log⁡58 \log_2 3x + \log_5 8 directly = invalid operation! You cannot combine logarithms with different bases using addition or multiplication rules. Always convert to the same base using change of base formula 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 can't I just add logarithms with different bases?

+

Logarithm rules like product and quotient rules only work when all logarithms have the same base. Different bases require the change of base formula first: log⁡ab=log⁡cblog⁡ca \log_a b = \frac{\log_c b}{\log_c a}

How do I know which base to convert to?

+

Choose the base that appears most frequently or simplifies the calculation. In this problem, converting to base 2 eliminated the fraction log⁡22log⁡25 \frac{\log_2 2}{\log_2 5} since log⁡22=1 \log_2 2 = 1 !

What does the cube root in the answer mean?

+

The cube root 2a2273 \sqrt[3]{\frac{2a^2}{27}} comes from raising both sides to power 13 \frac{1}{3} when we had log⁡23x=13log⁡2(2a2) \log_2 3x = \frac{1}{3}\log_2(2a^2) . This is equivalent to 3x=(2a2)1/3 3x = (2a^2)^{1/3} .

How can I verify this complex answer?

+

Substitute x=2a2273 x = \sqrt[3]{\frac{2a^2}{27}} back into the left side. Calculate log⁡2(3⋅2a2273)×log⁡58 \log_2(3 \cdot \sqrt[3]{\frac{2a^2}{27}}) \times \log_5 8 and check if it equals log⁡5(2a2) \log_5(2a^2) .

Why did we get 27 in the denominator?

+

When we solved 3x=(2a2)1/3 3x = (2a^2)^{1/3} for x, we got x=(2a2)1/33 x = \frac{(2a^2)^{1/3}}{3} . To write this as one cube root: 13=1333=127 \frac{1}{3} = \frac{1^3}{3^3} = \frac{1}{27} under the cube root.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Rules of Logarithms questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime

More Questions

Click on any question to see the complete solution with step-by-step explanations

Rules of Logarithms

Solve Complex Logarithmic Equation: log₅9(log₃4x + log₃(4x+1)) = 2(log₅4a³ - log₅2a)
Click to solve
Solve: log₉(e³) × (log₂24 - log₂8) × ln(8×2) | Complete Logarithm Problem
Click to solve

Multiplication of Logarithms

Solve Complex Logarithm Equation: Product of Log_a(x), Log_b(y), and Log_c(2)
Click to solve
Solve Mixed Logarithmic Equation: ln(8x) and log₇(e²) Problem
Click to solve

Rules of Logarithms Combined

Calculate ln(4) with Mixed Base-2 and Base-7 Logarithmic Expression
Click to solve
Solve the Logarithmic Equation: ln x + ln(x+1) - ln2 = 3
Click to solve

The Sum of Logarithms

Solve log₄(9x) + log₄(x+4) - log₄(3) = ln(2e) + ln(1/2e): Finding X
Click to solve
Solve the Complex Logarithmic Equation: log₅x + log₅(x+2) + log₂5 - log₂2.5 = log₃7 × log₇9
Click to solve