Question

4. Evaluate log x + log( x - 1) = log(3x + 12) (a) -2,6 (b) -2,-6 (c) 3,4​

Answer 1

Given :

$\sf\log(x)+\log(x+1)=\log(3x+12)$

To be found :

the value of x

Solution

We have to find the value of x

$\sf\log(x)+\log(x+1)=\log(3x+12)$

We know that

$\sf\log(a)+\log(b)=\log(ab)$ , then

$\sf\log[x(x-1)]=\log(3x+12)$

$\sf\log[x^2-x]=\log(3x+12)$

Now on comparing on sides

x²- x = 3x + 12

x²- 4x - 12 = 0

By middle term splitting

x² - 6x + 2x - 12 = 0

x(x-6) + 2(x-6) = 0

(x-6)(x+2) = 0

x = 6, (-2)

Hence , correct option a)

Extra

• Basic rules

$\sf\:1)\log(a) + \log(b) = \log(ab)$

$\sf\:2)\log( \frac{a}{b} ) = \log(a) - \log(b)$

$\sf\:3)\log(a) {}^{n} = nlog(a)$

$\sf\:4)\log_{a}(a) = 1$

$\sf\:5)\log_{a}(1)=0$

$\sf\:6)\log_{a{}^{n}}x=\dfrac{1}{n}\times\log_{a}x$