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Solve the following equation for x

Solve the equation for x to three significant figures:
5^{2x} - 6(5^x)=0
I manage to get it to (5^{2x} - 6 * 5^x)+6*5^x = 0 + 6*5^x
so I end up with 5^{2x} = 6(5^x) but that's as far as I get.
Original post by Looneytunes76
Solve the equation for x to three significant figures:
5^{2x} - 6(5^x)=0
I manage to get it to (5^{2x} - 6 * 5^x)+6*5^x = 0 + 6*5^x
so I end up with 5^{2x} = 6(5^x) but that's as far as I get.


Do you know that 52x=(5x)25^{2x} = (5^x)^2

This is a quadratic

Could you solve

y26y=0y^2 - 6y = 0
You can represent 5^x as Y
Original post by TenOfThem
Do you know that 52x=(5x)25^{2x} = (5^x)^2

This is a quadratic

Could you solve

y26y=0y^2 - 6y = 0

would it be y = 6 squared rooted therefore y =2.45
Original post by Looneytunes76
would it be y = 6 squared therefore y =2.45


Nope

y26y=0y^2-6y=0

gives

y(y6)=0y(y-6)=0
Original post by TenOfThem
Nope

y26y=0y^2-6y=0

gives

y(y6)=0y(y-6)=0


sorry went off on a tangent there, so you've factorised it, still not sure about my question though I apologise for being so slow on the up take. 52x = 6 * 5x

however an example was very helpful last time. How can 5^x be used as y and would it be 6 (y)
(edited 9 years ago)
Original post by Looneytunes76
sorry went off on a tangent there, so you've factorised it, still not sure about my question though I apologise for being so slow on the up take. 52x = 6 * 5x

however an example was very helpful last time.


(5x)265x=0(5^x)^2 - 6*5^x = 0 is almost an identical question

But, an example :smile:

32x73x+10=03^{2x} - 7*3^x + 10 = 0

(3x5)(3x2)=0(3^x - 5)(3^x - 2) = 0

3x=53^x = 5 or 3x=23^x = 2


Then you take logs like we did in the previous question
Original post by TenOfThem
(5x)265x=0(5^x)^2 - 6*5^x = 0 is almost an identical question

But, an example :smile:

32x73x+10=03^{2x} - 7*3^x + 10 = 0

(3x5)(3x2)=0(3^x - 5)(3^x - 2) = 0

3x=53^x = 5 or 3x=23^x = 2


Then you take logs like we did in the previous question

-5-2 =-7 but also -5*-2 = +10 however in my equation how do I work out the factorisation as I don't have a number to represent the +10.
I don't know why everyone is so keen on solving quadratic equations? It seems simpler to me to apply logs directly. For an example with changed values:

7^(3x)=4*7^x
=>
log_7(7^(3x)) = log_7(4*7^x)
=>
log_7(7^(3x)) = log_7(4) + log_7(7^x)
=>
3x = log_7(4) + x
=>
2x = log_7(4)
=>
x = log_7(4)/2
Original post by Looneytunes76
-5-2 =-7 but also -5*-2 = +10 however in my equation how do I work out the factorisation as I don't have a number to represent the +10.


I already factorised yours (well the equivalent y version) for you
Original post by fat_hampster
I don't know why everyone is so keen on solving quadratic equations? It seems simpler to me to apply logs directly. For an example with changed values:

7^(3x)=4*7^x
=>
log_7(7^(3x)) = log_7(4*7^x)
=>
log_7(7^(3x)) = log_7(4) + log_7(7^x)
=>
3x = log_7(4) + x
=>
2x = log_7(4)
=>
x = log_7(4)/2


For 2 reasons

The OP is not very confident with Logs so I did not wish to introduce logs with a different base

In a more general version of this question (such as the example I gave) the OP will need to recognise the quadratic format
Original post by TenOfThem
I already factorised yours (well the equivalent y version) for you
I am completely lost. Would it be
52x-6(5x)=0
52x = 6*5x
log5(2x)=log5(6)+log5(5x)
2x-x=log5(6)+x-x
x=log5(6)
is this correct.
Original post by Looneytunes76
I am completely lost. Would it be
52x-6(5x)=0
52x = 6*5x
log5(2x)=log5(6)+log5(5x)
2x-x=log5(6)+x-x
x=log5(6)
is this correct.


That is a different method but, yes
Original post by fat_hampster
I don't know why everyone is so keen on solving quadratic equations? It seems simpler to me to apply logs directly. For an example with changed values:

7^(3x)=4*7^x
=>
log_7(7^(3x)) = log_7(4*7^x)
=>
log_7(7^(3x)) = log_7(4) + log_7(7^x)
=>
3x = log_7(4) + x
=>
2x = log_7(4)
=>
x = log_7(4)/2
Thanks for your help all the examples have helped me to answer the question.
Original post by TenOfThem
That is a different method but, yes


I got x=1.113 and once again would like to thank you, for all your help.

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