The two questions are as follow and the image attached shows all my steps towards attempting to solve them:
a) $1+ \log y = \log (y+3)$: I am missing something since my steps do not make sense. IF I collect the $\log y$ terms, they cancel each other out when I bring it to the other side to isolate for $y$.
b)$\log_2 (x - 3) + \log_2 (x + 5) − \log_2 (x + 15) =0$: I managed to get two solutions $x= -5$ and $x = 4$, but when I input those values into the original equation, my answer does not equal $0$. I reject $-5$ as an erroneous root because it results in negative values for log. So with $x=4$ left, I get: \begin{align*} \log_2(1) + \log_2(9) - \log_2(19) & = \frac{\log_2(9)}{\log_2(19)} && \text{used product and quotient rules}\\ & = 0.7462285999 \end{align*} so l.s. does not equal r.s.
I appreciate any help or tips you may offer, thank you.
a) I assume the base is 10
$1+\log(y) = \log(y+3)$
$1=\log(y+3) - \log(y)$ --> $1=\log(\frac{y+3}y)$
$10 = \frac{y+3}y$ ----> $10 = 1 + \frac3y$ ----> $y=\frac3{10-1}=\frac13$
What you're missing: Incorrect application of product rule for logarithms.
b) I assume the base is 2.
$\log_2(x-3) + \log_2(x+5) - \log_2(x+15)=0$. I'll move over the negative log to the right side.
$\log_2(x-3) + \log_2(x+5) = \log_2(x+15)$. Combine the left side.
$\log_2(x^2+2x-15) = \log_2(x+15)$. 2^(everything) comes next.
$x^2+2x-15=x+15$ ----> $x^2+x-30=0$ --> $(x+6)(x-5)=0$
$-6$ is invalid because you cannot have negative logarithms. Final answer: $5$.
Check:
$\log_2(5-3) + \log_2(5+5) - \log_2(5+15)=^?=0$
$\log_2(2) + \log_2(10) - \log_2(20)=^?=0$
$1+\log_2(\frac12)=^?=0$
$1-1=0$
What you were missing: A little mistake when you wrote down $x^2+2x-5$ instead of $x^2+2x-15$
I hope you're happy with everything!