Logarithms/exponential equations Watch

Mysticmeg
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#1
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#1
I'm stuck on part of a question:

9) Solve these questions without using a calculator
(a) 6^x = 6^(2x+1)

(b) 4^(4x+6)=4^(7x-3)

(c) 2^(4x+6)=4^(7x-3)

I can do part (a):
6^x= (6^x)^2(6^1)
6y^2 - y
y(6y-1)
y=0, y=1/6
6^x = 1/6
so x=-1

But I can't figure out how to do part (b) or (c)
My attempt at (B):
(4^x)^4(4^6)= (4^x)^7(4^-3)
(4^6)Y^4=y^7(4^-3)
(3096)y^4=(1/64)y^7
this looks way too complicated.
Also how do you do part (c) because there are two different bases?
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lazy_fish
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#2
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(Original post by Mysticmeg)
...
 x^a = x^b \Rightarrow a = b

 4 = 2^2
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bethanoconnor
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If the base numbers are the same then just let the powers equal each other and then solve for x!
It's actually way more simple than how you did part (a)


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davros
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#4
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(Original post by Mysticmeg)
I'm stuck on part of a question:

9) Solve these questions without using a calculator
(a) 6^x = 6^(2x+1)

(b) 4^(4x+6)=4^(7x-3)

(c) 2^(4x+6)=4^(7x-3)

I can do part (a):
6^x= (6^x)^2(6^1)
6y^2 - y
y(6y-1)
y=0, y=1/6
6^x = 1/6
so x=-1

But I can't figure out how to do part (b) or (c)
My attempt at (B):
(4^x)^4(4^6)= (4^x)^7(4^-3)
(4^6)Y^4=y^7(4^-3)
(3096)y^4=(1/64)y^7
this looks way too complicated.
Also how do you do part (c) because there are two different bases?
You're making heavy weather of this!

If 6^a = 6^b then a = b so for part (a) you can write x = 2x + 1 and solve giving x = -1 as you have correctly found.

Now try the same method for part (b).
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Zacken
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(Original post by lazy_fish)
 x^a = x^b \Rightarrow a = b

 4 = 2^2
I was just a little curious as to what conditions should be placed on x for the implication you mention to hold. Definitely, we have that x \neq -1, 0, 1? Any other conditions?
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Mysticmeg
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(Original post by lazy_fish)
 x^a = x^b \Rightarrow a = b

 4 = 2^2
So for (c) would I
4x + 6 = 7x - 3

square root the left side or square the left side?
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davros
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#7
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(Original post by Mysticmeg)
So for (c) would I
4x + 6 = 7x - 3

square root the left side or square the left side?
Don't square root either side!!

Express both sides as powers of 2 and then follow the same process as for parts (a) and (b).
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Graph
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Rewrite 4 as 2^2, and then remember that (x^a)^b = x^{ab}. Once the bases are the same, you can simply equate the powers then solve for x
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Mysticmeg
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#9
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Thanks
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TeeEm
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(Original post by lazy_fish)
 x^a = x^b \Rightarrow a = b

 4 = 2^2
I owed you a rep...
I have a vague recollection of your "username picture"
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