Question

(Will award brainiest) Need help finding errors in this problem. Any help is appreciated!

Answer 1

Answer:

See below

Step-by-step explanation:

We would like to solve and the figure out mistakes of the given logarithmic question ,

$\longrightarrow ln (x) + ln(x-3)= ln(10)$

Prerequisites : Properties of logarithms

• $log_a(m) + log_a(n) = log_a(mn)$

• $log_a(m) - log_a(n)=log_a\dfrac{m}{n}$

Here $ln$ means $log_e$ , where $e$ is Euler's Number . So in the given equation all logs have base " e " . Using first property stated above , we have ;

$\longrightarrow ln ( x(x-3)) = ln(10)$

Simplify, inside the brackets ;

$\longrightarrow ln( x^2-3x)=ln(10)$

Now on comparing , we have ;

$\longrightarrow x^2-3x =10$

Subtract 10 on both sides ,

$\longrightarrow x^2-3x-10=0$

Solve the quadratic equation by factorising ,

$\longrightarrow x^2-5x +2x -10=0$

$\longrightarrow x(x-5)+2(x-5)=0$

$\longrightarrow (x-5)(x+2)=0$

Equate each factor with 0 ,

$\longrightarrow x = 5,-2$

As we can't put negative values in logarithm,

$\longrightarrow \underline{\underline{\boldsymbol{x = 5}}}{}$

Mistakes made by Aarnie :

Aarnie made mistake in the very first line of his answer. Instead of multiplying $x$ and $x-3$ he added them up . Possibly he might have used $log_a(m) + log_a(n) = log_a(m+n)$ , which is totally wrong .

(The correct way of solving is shown above .)