Question

log(x+1) to base 3 - 1 = 3 + log(x-1) to base 3​

Answer 1

Question :

• $\sf log_{3}(x + 1) - 1 = 3 + log_{3}(x - 1)$

Solution :

$\sf log_{3}(x + 1) - 1 = 3 + log_{3}(x - 1)$

$\implies\sf log_{3}(x + 1) - log_{3}(x - 1) = 3 + 1$

$\implies \sf log_{3}( \frac{x + 1}{x - 1} ) = 4$

• we know that

$\sf log_{x}(y) = a$

$\sf \implies a {}^{x} = y$

• Therefore :

$\sf4 {}^{3} = \dfrac{x + 1}{x - 1}$

$\implies \sf64 = \dfrac{x + 1}{x - 1}$

$\implies \sf64(x - 1) = x + 1$

$\sf \implies64x - 64 = x + 1$

$\sf \implies64x - x = 1 + 64$

$\sf \implies63x = 65$

$\sf \implies x = \dfrac{65}{63}$

Formulas Used :

$\sf1) log(x) - log(y) = log( \frac{x}{y} )$

$\sf2) \sf2) log(x) + log(y) = log(xy)$