Question

Log10 x +log 10(x−1)=log 10 (3x+12)

Answer 1

Given

• Log Equation
• $\sf{Log_{10}^x + Log_10^{x-1}=Log_10^{3x+12}}$

To find

• Value of x

Solution

$\sf{Log_{10}^x + Log_{10}^{x-1}=Log_10^{3x+12}}$

$\implies\sf{Log_{10}^x + Log_{10}^{x-1}=Log_{10}^{3x+12}}$

Same bases 10

$\implies\sf{Log_{10}^{(x)(x-1)}=Log_{10}^{3x+12}}$

Both side comparing the bases

=>x²-x=3x+12

=>x²-x-3x-12=0

=>x²-4x-12=0

=>x²-6x+2x-12=0

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

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

=>X= -2,6

Here -2 is not satisfied when we put in the Equation

Therefore X = 6