How do you solve $log x + log (x + 21) = 2$ $log x + log(x+21) =2$ ?

Question

13702 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-20T12:26:55

$x = 4$ $x=4$

Here,

$log x + log (x + 21) = 2$ $logx+log(x+21)=2$

We assume that the common base of the log as 10.

${log}_{10} x + {log}_{10} (x + 21) = 2$ $log_10x+log_10(x+21)=2$

$=>log_10(x*(x+21))=2to[becauselogM+logN=log(MN)$

$=>log_10(x^2+21x)=2$

$=>x^2+21x=10^2to[becauseX=log_aY<=>Y=a^X]$

$=>x^2+21x-100=0$

$=>x^2+25x-4x-100=0$

$=>x(x+25)-4(x+25)=0$

$=>(x-4)(x+25)=0$

$=>x=4 or x=-25$

But for $x=-25$ log x is not defined.

Hence $x=4$

Check :

$LHS=log_10 4+log_10 (4+21)=log_10 4+log_10 25$

$:.LHS=log_10(4xx25)=log_10 100=log_10 10^2=2=RHS$

How do you solve log x + log(x+21) =2logx+log(x+21)=2log x + log(x+21) =2?