# Solving two logarithmic questions: (image attached)

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.

• Please read this tutorial on how to typeset mathematics on this site. Commented May 31, 2018 at 10:13
• Oh okay, thank you for the link. Commented May 31, 2018 at 12:56

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!

• Hello Christopher, Thank you for your help. Surprising that I overlooked that mistake! For the first solution (a) The last line: how did you get '10 = 1 + 3/y'. I doubt this is what you meant but my guess was that you had the y's cancelled each other to get 1, although the y in the denominator should be cancelled too so I don't think this is what you meant. And following this step, to isolate the 'y' from the denominator, you multiply the 10 to get 10y, and then divide the 10 on both sides, in which case I got y = (1 + 3) / 10. How did the 1 end up in the denominator? Commented May 31, 2018 at 12:57
• @user554754 Because I assumed the base was 10, I took the entire expression and raised 10 to their powers: $a=b$ --> $10^a=10^b$ is what I was doing. When you have $\frac{y+3}y$, you can seperate the addends: $\frac{y}y + \frac3y$. Then I just maneuvered the numbers to solve for $y$. Hope this helps! You're welcome! Commented Jun 1, 2018 at 13:58