Question

NO LINKS!!

Solve each equation. Be sure to check for extraneous solutions. Part 2
k. log₇(x + 12) + log₇(x + 1) = log₇(42)

l . log₄(x + 3) + 1 = 2log₄(x)

Answer 1

k) x = 2

l) x = 6

Explanation:

`\boxed{\begin{minipage}{8cm}\underline{Log laws}\\\\Product law:\quad\:$\log_axy=\log_ax + \log_ay$\\\\Power law:\quad\;\;\;\:$\log_ax^n=n\log_ax$\\\\Equality Law:\; If $\log_ax=\log_ay$ then $x=y$\\\\Same base as number:\;\;\;$\log_aa=1$\\$\end{minipage}}`

Part k

`\begin{aligned}\log_7(x + 12) + \log_7(x + 1) &= \log_7(42)\\\log_7\left[(x + 12)(x+1)\right]&= \log_7(42)\\(x + 12)(x+1)&=42\\x^2+13x+12&=42\\x^2+13x-30&=0\\x^2+15x-2x-30&=0\\x(x+15)-2(x+15)&=0\\(x-2)(x+15)&=0\\\\\implies x&=2\\\implies x&=-15\end{aligned}`

As logs of negative numbers cannot be taken, x = -15 is an extraneous solution. Therefore, the only valid solution is x = 2.

Part l

`\begin{aligned}\log_4(x + 3) + 1 &= 2\log_4(x)\\\log_4(x + 3) + \log_44 &= \log_4(x^2)\\\log_4\left[4(x + 3)\right] &= \log_4(x^2)\\4(x + 3)&=x^2\\4x+12&=x^2\\x^2-4x-12&=0\\x^2-6x+2x-12&=0\\x(x-6)+2(x-6)&=0\\(x+2)(x-6)&=0\\\\\implies x&=-2\\\implies x&=6\end{aligned}`

As logs of negative numbers cannot be taken, x = -2 is an extraneous solution. Therefore, the only valid solution is x = 6.