Question

log3 (2x +1) + log3 (x + 8) = 3

Answer 1

$log3 ^(^2^x^+^1)^+ log3 ^(^x^+^8) = 3$

$log3 ^ \frac{(^2^x^+^1)}{ (^x^+^8) } = 3$

$\frac{(^2^x^+^1)}{ (^x^+^8) } =3^ 3$

$\frac{(^2^x^+^1)}{ (^x^+^8) } =27$

$(2x + 1) = 27x + 216$


$- 25x = 215$

$x =- \frac{ 215}{25}$

$x = - ( \frac{43}{5} )$