Solve: log₈(x³)/log₈(x^1.5) + log₇(x⁵)/log₄₉(x) Logarithmic Expression

Q: Why does the $ \log_8 x $ cancel out in the first fraction?

When you apply the power rule, you get $ \frac{3\log_8 x}{1.5\log_8 x} $. Since both numerator and denominator have the same $ \log_8 x $ term, they cancel out completely, leaving just $ \frac{3}{1.5} = 2 $!

Q: How do I know that $ \log_{49} x = \frac{\log_7 x}{2} $ ?

Use the change of base formula: $ \log_{49} x = \frac{\log_7 x}{\log_7 49} $. Since $ 49 = 7^2 $, we have $ \log_7 49 = \log_7 7^2 = 2 $, giving us $ \frac{\log_7 x}{2} $.

Q: Why does the second part equal exactly 10?

After simplifying, you get $ \frac{2}{\log_7 x} \times 5\log_7 x $. The $ \log_7 x $ terms cancel out: $ 2 \times 5 = 10 $. It's that simple!

Q: Can I solve this without change of base formula?

Not easily! The change of base formula is essential here because it converts $ \log_{49} x $ to something involving $ \log_7 x $, which then cancels with the $ \log_7 x^5 $ term.

Q: What if I get confused by all the different bases?

Focus on one piece at a time! First, simplify the fraction with base 8. Then work on the base 7 and base 49 part separately. Look for patterns - notice that $ 49 = 7^2 $ connects the bases.

Logarithmic Properties with Change of Base

$\frac{\log_8x^3}{\log_8x^{1.5}}+\frac{1}{\log_{49}x}\times\log_7x^5=$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:06 We'll use the formula for logarithmic division

00:11 We'll get the log of the numerator in base of the denominator

00:16 We'll use this formula in our exercise

00:26 We'll raise the product to the numerator

00:41 We'll solve the logarithm

01:07 We'll put this solution in the exercise and continue solving

01:22 We'll use the formula for 1 divided by log

01:27 We'll get the inverse logarithm in base and number

01:32 We'll use this formula in our exercise

01:42 We'll use the formula for logarithmic multiplication, we'll switch between the numbers

01:57 We'll use this formula in our exercise

02:12 We'll calculate each logarithm separately and put it in the exercise

02:57 And this is the solution to the question

Step-by-step written solution

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Understand the problem

$\frac{\log_8x^3}{\log_8x^{1.5}}+\frac{1}{\log_{49}x}\times\log_7x^5=$

Step-by-step solution

To solve the given problem, we begin by simplifying each component of the expression.

Step 1: Simplify $\frac{\log_8x^3}{\log_8x^{1.5}}$ .
Applying the power rule of logarithms, we get:
$\log_8x^3 = 3 \log_8x$ , and $\log_8x^{1.5} = 1.5 \log_8x$ .
Thus, $\frac{3 \log_8x}{1.5 \log_8x} = \frac{3}{1.5} = 2$ .

Step 2: Simplify $\frac{1}{\log_{49}x} \times \log_7x^5$ .
First, notice that $\log_7x^5 = 5 \log_7x$ by the power rule.
Applying the change of base formula, $\log_{49}x = \frac{\log_7x}{\log_749} = \frac{\log_7x}{2}$ because $49 = 7^2$ .
This gives $\frac{1}{\log_{49}x} = \frac{2}{\log_7x}$ .
Therefore, $\frac{2}{\log_7x} \times 5 \log_7x = 2 \times 5 = 10$ .

Step 3: Combine the results from Step 1 and Step 2.
The simplified expression is $2 + 10 = 12$ .

Therefore, the solution to the problem is $12$ .

Final Answer

$12$

Key Points to Remember

Essential concepts to master this topic

Power Rule: $\log_a x^n = n \log_a x$ reduces complex exponents
Change of Base: $\log_{49} x = \frac{\log_7 x}{\log_7 49} = \frac{\log_7 x}{2}$ since $49 = 7^2$
Check: Each simplified term cancels variables: $\frac{3\log_8 x}{1.5\log_8 x} = 2$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to apply logarithm properties before simplifying
Don't leave expressions like $\frac{\log_8 x^3}{\log_8 x^{1.5}}$ unsimplified = messy fractions with variables! Without using the power rule first, you can't cancel terms and will get stuck with complicated expressions. Always apply $\log_a x^n = n \log_a x$ first to convert to simple coefficients.

Practice Quiz

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$\log_{10}3+\log_{10}4=$

FAQ

Everything you need to know about this question

Why does the $\log_8 x$ cancel out in the first fraction?

When you apply the power rule, you get $\frac{3\log_8 x}{1.5\log_8 x}$ . Since both numerator and denominator have the same $\log_8 x$ term, they cancel out completely, leaving just $\frac{3}{1.5} = 2$ !

How do I know that $\log_{49} x = \frac{\log_7 x}{2}$ ?

Use the change of base formula: $\log_{49} x = \frac{\log_7 x}{\log_7 49}$ . Since $49 = 7^2$ , we have $\log_7 49 = \log_7 7^2 = 2$ , giving us $\frac{\log_7 x}{2}$ .

Why does the second part equal exactly 10?

After simplifying, you get $\frac{2}{\log_7 x} \times 5\log_7 x$ . The $\log_7 x$ terms cancel out: $2 \times 5 = 10$ . It's that simple!

Can I solve this without change of base formula?

Not easily! The change of base formula is essential here because it converts $\log_{49} x$ to something involving $\log_7 x$ , which then cancels with the $\log_7 x^5$ term.

What if I get confused by all the different bases?

Focus on one piece at a time! First, simplify the fraction with base 8. Then work on the base 7 and base 49 part separately. Look for patterns - notice that $49 = 7^2$ connects the bases.

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Solve: log₈(x³)/log₈(x^1.5) + log₇(x⁵)/log₄₉(x) Logarithmic Expression

Logarithmic Properties with Change of Base

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why does the log⁡8x \log_8 x log8​x cancel out in the first fraction?

How do I know that log⁡49x=log⁡7x2 \log_{49} x = \frac{\log_7 x}{2} log49​x=2log7​x​?

Why does the second part equal exactly 10?

Can I solve this without change of base formula?

What if I get confused by all the different bases?

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Why does the $\log_8 x$ cancel out in the first fraction?

How do I know that $\log_{49} x = \frac{\log_7 x}{2}$ ?