If _a ≤ft( {19³{[4]{3^5} × 7² ) = 3_a 19 - 2_a 7 - x then which one of the following is true?

Correct Answer is : ax=35/4a^x = 3^{5/4}ax=35/4

Correct Answer is : ax=35/4a^x = 3^{5/4}ax=35/4

Test Index

Logarithms Practice Test 2

Question : 2 of 39

Marks: +1, -0

\log_a \left( \frac{19^3}{\sqrt[4]{3^5} \times 7^2} \right) = 3\log_a 19 - 2\log_a 7 - x

then which one of the following is true?

Solution:

Concept:

Use

\log_a\frac{m}{n}=\log_a m-\log_a n

\log_a(mn)=\log_a m+\log_a n

, and

\log_a m^n=n\log_a m

to compare both sides.

Explanation:

Write

\sqrt[4]{3^5}=3^{5/4}

Apply quotient and product rules to the left side:

\log_a\left(\frac{19^3}{3^{5/4}\cdot7^2}\right)=3\log_a19-\frac54\log_a3-2\log_a7

Equate it to the right side:

3\log_a19-\frac54\log_a3-2\log_a7=3\log_a19-2\log_a7-x

Cancel

3\log_a19

and

-2\log_a7

from both sides:

-\frac54\log_a3=-x

So,

x=\frac54\log_a3=\log_a 3^{5/4}

Convert logarithmic form to exponential form:

a^x=3^{5/4}

Answer:

Option B:

a^x=3^{5/4}

Go to Question:

Prev Question

Next Question