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Can you solve this without trial and error?

5^x +4^x =8

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x must be less than 1
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Avatar for mariejuana
mariejuana
OP
10
Original post by the bear
x must be less than 1

I know, but can I figure the solution without trial and error?
The exact value
x is bigger than 0
Original post by mariejuana
I know, but can I figure the solution without trial and error?
The exact value


Change the bases.
Hay ho
On
Eb
Reply 8
Avatar for mariejuana
mariejuana
OP
10
Original post by RDKGames
Change the bases.


I did and it gave me a weird answer.
Rty
Original post by RDKGames
What was the weird answer?


i changed 4^x t0 2^2x. changed 8 to 2^3
used logs to change 5 t0 a power of 2

2^x= 5
x=log5/log2
so 2^log5/log2 x + 2^2x= 2^3
Original post by mariejuana
i changed 4^x t0 2^2x. changed 8 to 2^3
used logs to change 5 t0 a power of 2

2^x= 5
x=log5/log2
so 2^log5/log2 x + 2^2x= 2^3


Sorry, ignore what I said as it doesn't apply here... long day :frown:

You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be xn+1=log⁡5(8−4xn)\displaystyle x_{n+1}=\log_5(8-4^{x_n}) and start with x1=12x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.
(edited 6 years ago)
Original post by RDKGames
Sorry, ignore what I said as it doesn't apply here... long day :frown:

You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be xn+1=log⁡5(8−4xn)x_{n+1}=\log_5(8-4^{x_n}) and start with x1=12x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.



Thank youuu!!
Original post by RDKGames
Sorry, ignore what I said as it doesn't apply here... long day :frown:

You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be xn+1=log⁡5(8−4xn)\displaystyle x_{n+1}=\log_5(8-4^{x_n}) and start with x1=12x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.




What about 5^x + 2*5^1-x =7 ?
Original post by mariejuana
What about 5^x + 2*5^1-x =7 ?


You can easily solve that and find an exact solution.
Original post by RDKGames
You can easily solve that and find an exact solution.


how ? xD
Original post by mariejuana
how ? xD


Multiply through by 5x5^x to get (5x)2+10=7(5x)(5^x)^2+10=7(5^x) which is a quadratic.
Original post by RDKGames
Multiply through by 5x5^x to get (5x)2+10=7(5x)(5^x)^2+10=7(5^x) which is a quadratic.



i don't understand it at allllllllllllll
Original post by mariejuana
i don't understand it at allllllllllllll


Ok, it's literally nothing more difficult than solving a quadratic. Let y=5xy=5^x then you have y2+10=7yy^2+10=7y which is the same as y2−7y+10=0y^2-7y+10=0 which I'm sure you can solve before convering yy back to 5x5^x
Original post by mariejuana
i don't understand it at allllllllllllll


i do. doesn#t matter i was beingstupud

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