Question

Algebraically solve the equation: log5 + log5(2 + 9) = 1.

Answer 1

Given:

`log_5x+log_5(2x+9)=1`

Using the law of logarithm,

`logM+logN=log(MN)`

Hence,

`\begin{gathered} log_5x+log_5(2x+9)=1 \\ log_5(x(2x+9))=1 \\ log_5(2x^2+9x)=1 \end{gathered}`

Also, applying the logarithm rule;

`\begin{gathered} If\text{ }log_ab=x,\text{ then} \\ a^x=b \end{gathered}`

Thus;

`\begin{gathered} log_5(2x^2+9x)=1 \\ 5^1=2x^2+9x \\ 5=2x^2+9x \\ Collecting\text{ all terms to one side;} \\ 0=2x^2+9x-5 \\ 2x^2+9x-5=0 \end{gathered}`

Solving the quadratic equation by factorization;

`\begin{gathered} 2x^2+10x-x-5=0 \\ 2x(x+5)-1(x+5)=0 \\ (2x-1)(x+5)=0 \\ 2x-1=0\text{ OR }x+5=0 \\ 2x=1\text{ OR }x=-5 \\ x=(1)/(2) \\ \\ \\ Thus; \\ x=(1)/(2)\text{ OR }x=-5 \end{gathered}`
`\begin{gathered} log_5((1)/(2))+log_5(2((1)/(2))+9)=1 \\ log_5((1)/(2))+log_5(10)=1 \\ log_5((1)/(2)*10)=1 \\ log_55=1 \\ Hence,\text{ }x=(1)/(2)\text{ is a TRUE solution} \end{gathered}`

Also,

`\begin{gathered} log_5(-5)+log_5(2(-5)+9)=1 \\ log_5(-5)+log_5(-1)=1 \\ log_5(-5*-1)=1 \\ log_55=1 \\ Hence,\text{ }x=-5\text{ is also a TRUE solution} \end{gathered}`

Therefore, the solution to the equation is;

`x=(1)/(2),x=-5`