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Logarithms help!

How do I do this question-
Solve for x: 2^(2x) -2^(x) -6 =0
(edited 10 years ago)
Reply 1
Avatar for TenOfThem
TenOfThem
18
Original post by bobbricks
How do I do this question-
Solve for x: 2^(2x) -2^(x) -6 =0


Recognise that it is a quadratic
Original post by bobbricks
How do I do this question-
Solve for x: 2^(2x) -2^(x) -6 =0


Note that 22x=(2x)2Note \ that \ 2^{2x} = \left( 2^{x} \right)^2

∴(2x)2−2x−6=0\therefore \left( 2^{x} \right)^2 - 2^{x} - 6 = 0

Does this look familiar?
(edited 10 years ago)
Reply 3
Avatar for bobbricks
bobbricks
OP
19
Original post by Khallil
Note that 22x=(2x)2Note \ that \ 2^{2x} = \left( 2^{x} \right)^2

∴(2x)2−2x−6=0\therefore \left( 2^{x} \right)^2 - 2^{x} - 6 = 0

Does this look familiar?


So it's a quadratic?
What do I do after that- do I use the quadratic formula?
Original post by bobbricks
So it's a quadratic?
What do I do after that- do I use the quadratic formula?


Just use inspection.
Reply 5
Avatar for Munrot07
Munrot07
18
Say that 2^x = y and then substitute and solve for y
Original post by bobbricks
So it's a quadratic?
What do I do after that- do I use the quadratic formula?


That's definitely an option. You could also factorise the expression in terms of 2x2^{x} or even complete the square.

I personally find it much easier to do any of the above if I use a substitution as it is easier on my eyes. For example:

Spoiler


Then again, the spoiler is totally useless if you're comfortable with factorising the original expression in terms of 2x2^x
Reply 7
Avatar for bobbricks
bobbricks
OP
19
Original post by Khallil
That's definitely an option. You could also factorise the expression in terms of 2x2^{x} or even complete the square.

I personally find it much easier to do any of the above if I use a substitution as it is easier on my eyes. For example:

Spoiler


Then again, the spoiler is totally useless if you're comfortable with factorising the original expression in terms of 2x2^x

I've factorised it (by letting y=2^x) and have got y to be 3 or -2. Therefore, I have two equations to solve- 2^x=3 and 2^x=-2. However, I seem to be getting the wrong answer when I use the logarithm function on the calculator...Have I done something wrong?
Reply 8
Avatar for krisshP
krisshP
14
Original post by bobbricks
I've factorised it (by letting y=2^x) and have got y to be 3 or -2. Therefore, I have two equations to solve- 2^x=3 and 2^x=-2. However, I seem to be getting the wrong answer when I use the logarithm function on the calculator...Have I done something wrong?

From the spoiler in the quote,
(u-3)(u+2)=0

u=3 or u=-2

As we let u=2^x

3=2^x or -2=2^x

Second solution is invalid, leaving the first solution of
2^x =3

So you're stuck here. right?
Take log of both sides
log(2)^x =log3
bring down the power from a log rule, i think it's called the power rule.
xlog2 =log3
x=(log3)/(log2)
Just enter that on your calculator and get x :smile:
Original post by bobbricks
I've factorised it (by letting y=2^x) and have got y to be 3 or -2. Therefore, I have two equations to solve- 2^x=3 and 2^x=-2. However, I seem to be getting the wrong answer when I use the logarithm function on the calculator...Have I done something wrong?


(2x)2−2x−6=0\left( 2^{x} \right)^2 - 2^{x} - 6 = 0

Let y=2x ⇒ ∴ y2−y−6=0 ⇒ y=1±252Let \ y = 2^{x} \ \Rightarrow \ \therefore \ y^2 - y - 6 = 0 \ \Rightarrow \ y = \dfrac{1 \pm \sqrt{25}}{2}

∴ y=−2,3 ⇒2x=−2 and 2x=3\therefore \ y = -2, 3 \ \Rightarrow 2^x = -2 \ and \ 2^x = 3

It should be quite clear that the first of these two isn't possible. A positive number to any power cannot yield a negative number.

For the second, it's helpful to take logarithms of base 2 on the equality.

This is because of the definition of a logarithm.

If log⁡ax=y, then ay=xIf \ \log_{a} x = y, \ then \ a^y = x


log⁡2(2x)=log⁡23\log_{2} (2^x) = \log_{2} 3

Using the power rule of logarithms yields:

xlog⁡22=log⁡23x \log_{2} 2 = \log_{2} 3

x=log⁡23x = \log_{2} 3
(edited 10 years ago)
Reply 10
Avatar for bobbricks
bobbricks
OP
19
Original post by krisshP
From the spoiler in the quote,
(u-3)(u+2)=0

u=3 or u=-2

As we let u=2^x

3=2^x or -2=2^x

Second solution is invalid, leaving the first solution of
2^x =3

So you're stuck here. right?
Take log of both sides
log(2)^x =log3
bring down the power from a log rule, i think it's called the power rule.
xlog2 =log3
x=(log3)/(log2)
Just enter that on your calculator and get x :smile:

That gives me an answer of 1.58... but the answers just say " log2(3) or 1 " -have I missed something (again)? :redface:

EDIT: I've just realised that the answer given is actually the exact form, so we went a few steps ahead...but that doesn't explain why the other answer is 1 (unless it's a typo) :redface: Thanks for the help so far though :biggrin:
(edited 10 years ago)
Reply 11
Avatar for bobbricks
bobbricks
OP
19
Original post by Khallil
(2x)2−2x−6=0\left( 2^{x} \right)^2 - 2^{x} - 6 = 0

Let y=2x ⇒ ∴ y2−y−6=0 ⇒ y=1±252Let \ y = 2^{x} \ \Rightarrow \ \therefore \ y^2 - y - 6 = 0 \ \Rightarrow \ y = \dfrac{1 \pm \sqrt{25}}{2}

∴ y=−2,3 ⇒2x=−2 and 2x=3\therefore \ y = -2, 3 \ \Rightarrow 2^x = -2 \ and \ 2^x = 3

It should be quite clear that the first of these two isn't possible. A positive number to any power cannot yield a negative number.

For the second, it's helpful to take logarithms of base 2 on the equality.



log⁡2(2x)=log⁡23\log_{2} (2^x) = \log_{2} 3

Using the power rule of logarithms yields:

xlog⁡22=log⁡23x \log_{2} 2 = \log_{2} 3

x=log⁡23x = \log_{2} 3

Thanks :smile:
I have 2 more questions though:
1) we haven't done much on logarithms in class so this might seem simple- why do we take base 2 instead of base 10 for instance?
2) The answers in the booklet states " log2(3) or 1 "- so is the latter still one of the solutions (if so, how have they got that) or do you think it's a typo/error?
Original post by bobbricks
Thanks :smile:
I have 2 more questions though:
1) we haven't done much on logarithms in class so this might seem simple- why do we take base 2 instead of base 10 for instance?
2) The answers in the booklet states " log2(3) or 1 "- so is the latter still one of the solutions (if so, how have they got that) or do you think it's a typo/error?


1. 1. \ It's perfectly acceptable to take logarithms of any base. Base 2 just seemed to be the most convenient one to take from my point of view since log⁡22=1\log_{2} 2 = 1 The generally accepted form of an answer like this is log(3)log(2)\frac{log(3)}{log(2)} since a base isn't specified and any base can be chosen.

2. 2. \ I'm not exactly sure how they got 1 as an alternative answer. The graph appears to have only one root at log(3)log(2)\frac{log(3)}{log(2)}

Here is the graph:
(edited 10 years ago)
Reply 13
Avatar for bobbricks
bobbricks
OP
19
Original post by Khallil
1. 1. \ It's perfectly acceptable to take logarithms of any base. Base 2 just seemed to be the most convenient one to take from my point of view since log⁡22=1\log_{2} 2 = 1 The generally accepted form of an answer like this is log(3)log(2)\frac{log(3)}{log(2)} since a base isn't specified and any base can be chosen.

2. 2. \ I'm not exactly sure how they got 1 as an alternative answer. The graph appears to have only one root at log(3)log(2)\frac{log(3)}{log(2)}

Here is the graph:


Ah, okay, thanks- that probably means it was just a type :smile:
No problem :wink:

Original post by bobbricks
Ah, okay, thanks- that probably means it was just a type :smile:


Reply 15
Avatar for bobbricks
bobbricks
OP
19
Original post by Khallil
No problem :wink:





Haha, well spotted :tongue:

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