Let a, b, c be real numbers,each greater than | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$.Let $x = \frac{2}{3} \log _{b} a$,$y = \frac{3}{5}

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(D) Given the equation: $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$.Let $x = \frac{2}{3} \log _{b} a$,$y = \frac{3}{5} \log _{c} b$,and $z = \frac{5}{2} \log _{a} c$.Note that $x \cdot y \cdot z = (\frac{2}{3} \log _{b} a) \cdot (\frac{3}{5} \log _{c} b) \cdot (\frac{5}{2} \log _{a} c) = (\frac{2}{3} \cdot \frac{3}{5} \cdot \frac{5}{2}) \cdot (\log _{b} a \cdot \log _{c} b \cdot \log _{a} c) = 1 \cdot 1 = 1$.By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$,for positive real numbers $x, y, z$ with $x+y+z=3$ and $xyz=1$,the equality holds only when $x=y=z=1$.Thus,$\frac{2}{3} \log _{b} a = 1 \Rightarrow \log _{b} a = \frac{3}{2}$.Given $b=9$,we have $\log _{9} a = \frac{3}{2}$.Therefore,$a = 9^{3/2} = (3^2)^{3/2} = 3^3 = 27$.

Let $a, b, c$ be real numbers,each greater than $1$,such that $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$. If the value of $b$ is $9$,then the value of $a$ must be

Similar Questions

If $a^x = b^y = c^z = d^w$,then the value of $x\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{w}\right)$ is

If $\sqrt{\log_3 x^{16}} + 9 \log_{27} \sqrt[3]{\frac{3}{x}} = 5$,then $x = \dots$.

Let $n$ be a positive integer such that $\log _2 \log _2 \log _2 \log _2 \log _2(n) < 0 < \log _2 \log _2 \log _2 \log _2(n)$. Let $l$ be the number of digits in the binary expansion of $n$. Then the minimum and the maximum possible values of $l$ are

If ${\log _{10}}x = y$,then ${\log _{1000}}{x^2}$ is equal to

$x = \log \left( \frac{1}{y} + \sqrt{1 + \frac{1}{y^2}} \right) \Rightarrow y$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module