Question

log (3x+2) base 10 +log(x-1) base 10 =1

Answer 1

Solution:

Given Equation:

$\rm \longrightarrow log_{10}(3x + 2) + log_{10}(x - 1) = 1$

$\rm \longrightarrow log_{10}(3x + 2)(x - 1) = 1$

$\rm \longrightarrow log_{10}(3 {x}^{2} - 3x + 2x - 2) = 1$

$\rm \longrightarrow log_{10}(3 {x}^{2} -x - 2) = 1$

$\rm \longrightarrow 3 {x}^{2} -x - 2 = {10}^{1}$

$\rm \longrightarrow 3 {x}^{2} -x - 12 = 0$

$\rm \longrightarrow x = \dfrac{1 \pm \sqrt{1 + 4 \times 3 \times 12} }{6}$

$\rm \longrightarrow x = \dfrac{1 \pm \sqrt{144} }{6}$

But x > 0 and therefore:

$\rm \longrightarrow x = \dfrac{1 + \sqrt{144} }{6}$

★ Which is our required answer.

Answer:

$\rm \hookrightarrow x = \dfrac{1 + \sqrt{144} }{6}$

Basic Concepts Used:

$\rm1. \: \: log(x) + log(y) + .. = log(xy...)$

$\rm2. \: \: log_{a}(x) = y \: or \: x = {a}^{y}$

Answer 2

Refer The Given Attachment.