Author Topic: Log graphs (Read 3941 times) Tweet Share

ngRISING · « **Reply #15 on:** September 27, 2009, 07:02:53 pm »

Quote from: Damo17 on September 27, 2009, 02:46:56 pm

Quote from: moshi on September 27, 2009, 02:28:18 pm
sketch y = log0.5(|2x - 2|) *base 0.5

x-intercept:
solve $log_{0.5}(2x-2)=0$ , so $x=1.5$
solve $log_{0.5}(2-2x)=0$ , so $x=0.5$

y-intercept: let $x=0$
$y=log_{0.5}(2)$

Asymptote: $x=1$

and sketch from there.

the x-axis asymtope is from the modulus in the brackets right. rusty on logs. :S
2x-2=0
x=1

Flaming_Arrow · « **Reply #16 on:** September 27, 2009, 07:04:46 pm »

Quote from: ngRISING on September 27, 2009, 07:02:53 pm

Quote from: Damo17 on September 27, 2009, 02:46:56 pm
Quote from: moshi on September 27, 2009, 02:28:18 pm
sketch y = log0.5(|2x - 2|) *base 0.5

x-intercept:
solve $log_{0.5}(2x-2)=0$ , so $x=1.5$
solve $log_{0.5}(2-2x)=0$ , so $x=0.5$

y-intercept: let $x=0$
$y=log_{0.5}(2)$

Asymptote: $x=1$

and sketch from there.

the x-axis asymtope is from the modulus in the brackets right. rusty on logs. :S
2x-2=0
x=1

correct

ngRISING · « **Reply #17 on:** September 27, 2009, 07:14:06 pm »

i drew the graph. i got it wrong sadly

can someone explain to me why the correct answers like that T_T>

TrueTears · « **Reply #18 on:** September 27, 2009, 07:16:49 pm »

Quote from: kamil9876 on September 27, 2009, 03:28:59 pm

generalisation:

$y=log_r u$

$r^y=u$

$(r^{-1})^{-y}=u$

$y=-log_{r^{-1}} (u)$

In fact you can also use dekoyl's change of base formula:

$log_a(b) = \frac{log_c(b)}{log_c(a)}$

and set $c=\frac{1}{a}$ which implies $a=c^{-1}$ , subbing this in gives:
$<br />log_a(b)=\frac{log_c(b)}{log_c(c^{-1})}$

$\implies log_a(b)=-log_c(b) where c=\frac{1}{a}$

Really can't put it anyway better than kamil did.

ngRISING · « **Reply #19 on:** September 27, 2009, 07:18:23 pm »

Quote from: Damo17 on September 27, 2009, 02:46:56 pm

Quote from: moshi on September 27, 2009, 02:28:18 pm
sketch y = log0.5(|2x - 2|) *base 0.5

x-intercept:
solve $log_{0.5}(2x-2)=0$ , so $x=1.5$
solve $log_{0.5}(2-2x)=0$ , so $x=0.5$

y-intercept: let $x=0$
$y=log_{0.5}(2)$

Asymptote: $x=1$

and sketch from there.

the y-int is -1. can someone also explain this ^^

TrueTears · « **Reply #20 on:** September 27, 2009, 07:20:45 pm »

QuantumJG · « **Reply #21 on:** September 27, 2009, 07:35:09 pm »

Quote from: ngRISING on September 27, 2009, 07:18:23 pm

Quote from: Damo17 on September 27, 2009, 02:46:56 pm
Quote from: moshi on September 27, 2009, 02:28:18 pm
sketch y = log0.5(|2x - 2|) *base 0.5

x-intercept:
solve $log_{0.5}(2x-2)=0$ , so $x=1.5$
solve $log_{0.5}(2-2x)=0$ , so $x=0.5$

y-intercept: let $x=0$
$y=log_{0.5}(2)$

Asymptote: $x=1$

and sketch from there.

the y-int is -1. can someone also explain this ^^

When graphing something, look and see if it makes logical sense!

If I have a fraction and put it to the power of 1 and then to the power of 2, which is larger?

Obviously the fomer is larger than the latter, but what you have said in your graph is that:

(1/2) < (1/2)^2

Search the forums now!

We have moved!

Author Topic: Log graphs (Read 3941 times) Tweet Share

ngRISING

Re: Log graphs

Flaming_Arrow

Re: Log graphs

ngRISING

Re: Log graphs

TrueTears

Re: Log graphs

ngRISING

Re: Log graphs

TrueTears

Re: Log graphs

QuantumJG

Re: Log graphs

Recent Posts

User:
Password: