Question

Find the largest value of x that satisfies:
log3(x^2)-log3(x+4)=4

Answer 1

`\bf \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( (x)/(y)\right)\implies \log_a(x)-\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^(log_a x)=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_3(x^2)-\log_3(x+4)=4\implies \log_3\left( \cfrac{x^2}{x+4} \right)=4`

`\bf 3^{\log_3\left( (x^2)/(x+4) \right)}=3^4\implies \cfrac{x^2}{x+4}=3^4\implies \cfrac{x^2}{x+4}=81\implies x^2=81x+324 \\\\\\ x^2-81x-324=0\implies x = \cfrac{81\pm √((-81)^2-4(1)(-324))}{2(1)} \\\\\\ x = \cfrac{81\pm √(6561+1296)}{2}\implies x = \cfrac{81\pm √(7857)}{2}\implies x = \begin{cases} x \approx -3.81986\\\\ x \approx 84.8199~~\checkmark \end{cases}`

Answer 2

Explanation:

log3(x^2)-log3(x+4)=4

Log 3(x^2/x+4)=4 -Combine the logs

x^2/x+4 = 3^4 - Use the relationship Logb(y)=x =>b^x=y

x^2/x+4 = 81

x^2= 81^(x+4)

x^2=81x+324

x^2-81x-324=0

Solving for x using the property x=-b-/+[tex]\sqrt{x} b^2-4ac]/ 2a

=(81 ±√(81)^2-4^11^324)/2

= (81±√6561-1296)/2

=(81±88.63)/2

= (81+88.63)/2

=84.81