How do you solve $ln (9 r + 1) = ln (r^{2} + 9)$ $ln(9r+1)=ln(r^2+9)$ ?

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Answer 1 · 2016-09-03T19:14:38

The Soln. is $r = 8, 1$ $r=8,1$ .

Since, $ln$ $ln$ function is $1 - 1$ $1-1$ ,

$ln (9 r + 1) = ln (r^{2} + 9)$ $ln(9r+1)=ln(r^2+9)$

$\Rightarrow 9 r + 1 = r^{2} + 9, or, r^{2} - 9 r + 8 = 0$ $rArr 9r+1=r^2+9, or, r^2-9r+8=0$

$\Rightarrow (r - 8) (r - 1) = 0$ $rArr (r-8)(r-1)=0$

$\Rightarrow r = 8, r = 1$ $rArr r=8, r=1$ .

Both these roots satisfy the given eqn. Hence,

The Soln. is $r = 8, 1$ $r=8,1$ .

How do you solve ln(9r+1)=ln(r^2+9)ln(9r+1)=ln(r2+9)ln(9r+1)=ln(r^2+9)?