Question

Please solve it, find value of x in equation

ln(x²+x) = ln(x²) + ln(x) ​

Answer 1

Given : $\sf\ln(x^2+x) = \ln(x^2) + \ln(x)$

To find : Value of $\sf x$ in the given logarithmic equation.

Solution:-

We have,

$\sf \implies \ln( {x}^{2} + x)= \ln( {x}^{2} ) + \ln(x)$

$\sf \implies \ln(x \cdot(x+ 1))= \ln( {x}^{2} ) + \ln(x)$

$\sf \implies \ln(x) + \ln(x+ 1)= \ln( {x}^{2} ) + \ln(x)$

$\sf \implies \ln(x) - \ln(x)+ \ln(x+ 1)= \ln( {x}^{2} )$

$\sf \implies \ln(x+ 1)= \ln( {x}^{2} )$

Exponentiating both sides will give us:

$\sf \implies e^{\ln(x+ 1)}= e^{\ln( {x}^{2} ) }$

Apply logarithmic property:

• $\boxed{ \tt\purple {a^ {\log_ab} = b}}$

$\sf \implies (x + 1)= {x}^{2}$

$\sf \implies 0= {x}^{2} - (x + 1)$

$\sf \implies 0= {x}^{2} - x - 1$

Use quadratic formula to solve this equation.

$\sf \implies x = \dfrac{ - ( - 1) \pm \sqrt{ {( - 1)}^{2} - 4(1)( - 1) } }{2(1)}$

$\sf \implies x = \dfrac{ 1\pm \sqrt{ 1 + 4 } }{2}$

$\boxed{\bf \pink{ \implies x = \dfrac{ 1\pm \sqrt{5} }{2} }}$

Hence this is the required solution of given equation.

Learn More:

• $\boxed{\log(a \cdot b) = \log(a) + \log(b)}$

• $\boxed{ \log\left( \dfrac{a}{b} \right) = \log(a) - \log(b)}$

• $\boxed{\ln(x) = \log_e(x)}$

• $\boxed{\log_a(b) = \dfrac{ \log_c(b)}{ \log_c(a)}}$

• $\boxed{a^{ \log_a(x)} = x}$