Question

Find the square of sum of roots if the equation -
log(base 3)x*log4x*log5x= log3x*log4x+log4x*log5x+log5x*log3x. (the no. 3, 4 and 5 are the bases of log x.
Thanks!

Answer 1

hope this helps you ☺☺

Answer 2

Answer:

Concept:

Solving Logarithm equations

An equation using the logarithm of an expression containing a variable is referred to as a logarithmic equation. Check to verify if you can write both sides of the equation as powers of the same number before attempting to solve an exponential equation.

Step-by-step explanation:

Given:

$\log _{3} x \cdot \log _{4} x \cdot \log _{5} x=\log _{3} x \cdot \log _{4} x+\log _{4} x \cdot \log _{5} x+\log _{5} x \cdot \log _{3} x$

Find:

Find the square of the sum of roots if the equation -$\log _{3} x \cdot \log _{4} x \cdot \log _{5} x=\log _{3} x \cdot \log _{4} x+\log _{4} x \cdot \log _{5} x+\log _{5} x \cdot \log _{3} x$

Solution:

We have

$\log _{3} x \cdot \log _{4} x \cdot \log _{5} x=\log _{3} x \cdot \log _{4} x+\log _{4} x \cdot \log _{5} x+\log _{5} x \cdot \log _{3} x$

          $For x=1 , \ both \ parts \ of \ the \ equation (1) \ vanish.\\\\Therefore,\ x=1 \ is \ root \ of \ the \ equation (1).\\\\For \ x \ not \ equal \ to \ 1\\\\ \log _{3} x \cdot \log _{4} x \cdot \log _{5} x=\log _{3} x \cdot \log _{4} x+\log _{4} x \cdot \log _{5} x+\log _{5} x \cdot \log _{3} x \\\\Divide both sides by \log _{3} x \cdot \log _{4} x \cdot \log _{5} x$

             $\\ &\Rightarrow 1=\frac{1}{\log _{5} x}+\frac{1}{\log _{3} x}+\frac{1}{\log _{4} x}\\\\=\log _{x} 5+\log _{x} 3+\log _{x} 4 \quad\left[\because \log _{a} b=\frac{1}{\log _{b} a}\right] \\\\&=\log _{x} 60 \quad[\because \log a+\log b+\log c=\log (a b c)] \\\\&\Rightarrow x=60\\\\Thus, x=\{1,60\}$

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