# Thread: [GCSE-O] Logarithms And Indices

1. ## [GCSE-O] Logarithms And Indices

1)
Given y2 - 1 = x2
Show that log(y-1)x + log(y+1)x = 2(log(y-1)x)(log(y+1)x)
Here are my steps
Please let me know if I make any mistakes

log(y-1)(y+1)[(y-1)(y+1)]= log(y-1)(y+1)x2
log(y-1)(y+1)[(y-1)(y+1)]= 2 log(y-1)(y+1)x

log(y-1)[(y-1)(y+1)] / log(y+1)[(y-1)(y+1)] = 2 ( log(y-1)(y+1)x )

log(y-1)(y+1) / log(y+1)(y-1) = 2 ( log(y-1)(y+1)x )

I got stuck here. I am not sure how to move forward from here.

2)
Find the value of x such that (2/x)logn4 - (3/x)logn9 = 0
x2 - 2y2 = 2x - 3y
Here are my steps

logn4 [ logn(2/x) ] = logn9 [ logn (3/x) ]
logn22 [ logn(2/x) ] = logn32 [ logn (3/x) ]
logn22 [ logn(2) - logn(x)] = logn32 [ logn(3) - logn(x) ]
logn22 logn(2) - logn22 logn(x) = logn32 logn(3)- logn32 logn(x)

I got stuck here and I am not sure how to move forward from here.

3)
If 2x3y = 3x4y = 6
show by taking logarithms to base 2 and eliminating log23, that
x2 - 2y2 = 2x - 3y
Here are my following steps
2x3y = 3x4y = 6
2x/3x = 4y/3y
(2/3)x = (4/3)y

x log2(2/3) = y log2(4/3)
x [ log2(2) - log2(3) ] = y [ log2(4) - log2(3) ]
x log2(2) - 2y log2(2) = x log2(3) - y log2(3)
x - 2y = log2(3) [(x-y)]
log2(3) = (x - y)/(x - 2y)

I know I did something wrong here but I just couldn't figure out.

Thank you very much. Hope someone can help me out! I will appreciate it. I don't need a full solution but I would like some hints on how should I carry on. Thank you!!

2. ## Re: [GCSE-O] Logarithms And Indices

Hello, mynameiscosine!

Here is #3.

$\displaystyle \text{3. Given: }\:2^x3^y \:=\:3^x4^y \:=\:6$

$\displaystyle \text{show by taking logs (base 2) and eliminating }\log_23$

. . $\displaystyle \text{that: }\:x^2-2y^2 \:=\:2x - 3y$

We have: .$\displaystyle 2^x3^y \:=\:6$

Take logs, base 2: .$\displaystyle \log_2(2^x3^y) \:=\:\log_2(6)$

n . . . . . . .$\displaystyle \log_2(2^x) + \log_2(3^y) \:=\:\log_2(2\cdot3)$

n . . . . . . . . $\displaystyle x\underbrace{\log_22}_1 + y\log_23 \:=\:\underbrace{\log_22}_1 + \log_23$

. . . . . . . . . . . . .$\displaystyle x + y\log_23 \:=\:1 + \log_23$

. . . . . . . . . . $\displaystyle y\log_23 - \log_23 \:=\:1 - x$

. . . . . . . . . . . . $\displaystyle (y-1)\log_23 \:=\:1-x$

. . . . . . . . . . . . . . . . . $\displaystyle \log_23 \:=\:\frac{1-x}{y-1}$ .[1]

We have: .$\displaystyle 3^x(2^2)^y \:=\:6 \quad\Rightarrow\quad 3^x2^{2y} \:=\:6$

Take logs, base 2: .$\displaystyle \log_2(3^x2^{2y}) \:=\:\log_2(6)$

n . . . . . . . $\displaystyle \log_2(3^x) + \log_2(2^{2y}) \:=\:\log_2(2\cdot 3)$

. . . . . . . . . $\displaystyle x\log_23 + 2y\underbrace{\log_22}_1 \:=\:\underbrace{\log_22}_1 + \log_23$

. . . . . . . . . . . $\displaystyle x\log_23 - \log_23 \:=\:1 - 2y$

n . . . . . . . . . . . $\displaystyle (x-1)\log_23 \:=\:1-2y$

. . . . . . . . . . . . . . . . . $\displaystyle \log_23 \:=\:\frac{1-2y}{x-1}$ .[2]

Equate [1] and [2]: .$\displaystyle \frac{1-x}{y-1} \:=\:\frac{1-2y}{x-1}$

. . . . . . . . . $\displaystyle (1-x)(x-1) \:=\:(y-1)(1-2y)$

. . . . . . . . . $\displaystyle x - 1 - x^2 + x \:=\: y - 2y^2 - 1 + 2y$

. . . . . . . . . . . . . $\displaystyle x^2 - 2y^2 \:=\: 2x - 3y$

3. ## Re: [GCSE-O] Logarithms And Indices

@Soroban

Thank you so much!
I am still working on the other 2 questions.

Thanks! (:

By the way, how do I do the fractions symbol? I can't seems to find it. Thanks!