Question

.Given logb^3 = 1/4 what is logb^9

Answer 1

Answer:

1/2

Step-by-step explanation:

Find the domain of the inequality: $\left \{ {{b > 0} \atop {b\neq 1}} \right.$

Solve the inequality: b > 0

Convert inequality to equation: b = 1

Solve the equation: b = 1

Convert equation to inequality: b ≠ 1

Find the intersection: 0 < b < 1 or b > 1

The domain of the inequality is: $\text{log}_b3=\frac{1}{4} ,0 < b, < b < 1$ or$b > 1$

Convert logarithm to exponential form: $b^{\frac{1}{4} }$ = 3, 0 <b<b1 or b > 1

Power on both sides: (b^1/4)^4=3^4, 0<b<1 or b > 1

Simplify using exponent rule (a^n)^m = a^nm:

b = 3^4, 0<b<1 or b > 1

Calculate the power:

b = 81 , 0<b<1 or b > 1

Find the intersection: b = 81

Substitute: $\text{log}_{81}9$

Factorize the argument: $\text{log}_{81} 3^2$

Express the logarithm of a power of an expression as the power times the logarithm of the expression: $2*\text{log}_{81} 3$

Factorize the base: $2*\text{log}x_{3^4} 3$

Rewrite the expression using the formula: $\text{log}_{a^n} b$

to $(\frac{1}{n} ) \text{log}_ab:$ $2*\frac{1}{4} *\text{log}_33$

Apply the power law of logarithm to simplify the expression:

$2*\frac{1}{4}$

Reduce the expression to the lowest form:

$\frac{1}{2}$

Answer: 1/2

Answer 2

Answer:

$\log_b9=\dfrac{1}{2}$

Step-by-step explanation:

$\textsf{Given}: \quad\log_b3=\dfrac{1}{4}$

$\textsf{To find }\log_b9, \textsf{ rewrite 9 as 3}^2:$

$\implies \log_b9=\log_b3^2$

$\textsf{Apply the Power log law}: \quad \log_ax^n=n\log_ax$

$\implies \log_b3^2= 2\log_b3$

$\begin{aligned}\textsf{If }\log_b3& =\dfrac{1}{4}\\\\ \implies 2 \log_b3 & =2 \times \dfrac{1}{4}\\\\ & = \dfrac{2}{4}\\\\ & = \dfrac{1}{2}\end{aligned}$

$\textsf{Therefore}: \quad\log_b9=\dfrac{1}{2}$