Linear equation from log equation

Question

Further mathematics is driving crazy at the moment as I prepare for a PHD in chem eng.

I've been working hard at the books but this one has caught me out. I basically need to derive a linear equation.

$ \ \log_2(x + 1) -1 \ = \ 2 \ \ \log_2 y \ $

I know that I need to raise two to the power of both sides of the second equation

$2^{\log_2 z}=z$ is the law I'm going to utilize

Further, the laws of exponentials should lead me to the linear equation I'm after.

I'm at $(x+1)^2 -1 = 2^2 y$ but can't tell if I'm completely lost.

Log base 2? Please format your equations using subscripts etc... $\log_2$ is $\log_2$. — jdods, May 06 '16 at 12:18
By $2 \log^2 z= z$ do you mean $2 (\log_2 z)^2=z$ ? In this case, it is not a law... It is true for $z=2$ and $z \approx 0.67$, that's all. — Jean Marie, May 06 '16 at 12:28
Sorry all I have rectified my post to reflect what I was trying to state — New Zealand's finest, May 06 '16 at 12:39

user5954246 · Accepted Answer · 2016-05-06T12:51:04.363

4

$$\log_2(x+1)-\log_2 2$$ because you can write 1 as $\log_2 2$

now principle of logarithms: $\log_ab-\log_ac=\log_a(b/c)$ $$\implies\log_2(\frac{x+1}{2})$$ and this equation is equal to you LHS $$\log_2(\frac{x+1}{2})=2\log_2y$$ Now RHS you can write as $\log_2y^2$ because $a\log b=\log b^a$

now since bases of LHS and RHS are equal. we can equate to$$\frac{x+1}{2}=y^2$$

edited May 06 '16 at 12:51

answered May 06 '16 at 12:45

user5954246

1,135

1

You have a typo. It is $\frac{x+\color{red}1}{2}=y^2$ – callculus42 May 06 '16 at 12:49
@callculus where? – user5954246 May 06 '16 at 12:50
The red one-I think. – callculus42 May 06 '16 at 12:51
aha! thanks @callculus – user5954246 May 06 '16 at 12:51
You´re welcome. Now I can upvote the answer. But the domains are $x>-1$ and $y>0$ – callculus42 May 06 '16 at 12:54

score 0 · Answer 2 · answered May 06 '16 at 12:54

0

Unfortunately I cannot comment, but the answer should be

y = root((x+1)/2)

Because if you try to put a negative value for y in the question, it won't work. With the answer of

y^2 = (x+1)/2

Negative y values do work, and that is not the same.

answered May 06 '16 at 12:54

Bradman175

193

Linear equation from log equation

2 Answers2