Turn on thread page Beta
 You are Here: Home >< Maths

# Can you solve this without trial and error? watch

1. 5^x +4^x =8
2. x must be less than 1
3. (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
4. x is bigger than 0
5. (Original post by mariejuana)
I know, but can I figure the solution without trial and error?
The exact value
Change the bases.
6. Hay ho
7. On
8. Eb
9. (Original post by RDKGames)
Change the bases.
I did and it gave me a weird answer.
10. Rty
11. (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
12. (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

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 and start with as an example. This also means what you've done makes no sense.
13. (Original post by RDKGames)
Sorry, ignore what I said as it doesn't apply here... long day

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 and start with as an example. This also means what you've done makes no sense.

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

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 and start with as an example. This also means what you've done makes no sense.

What about 5^x + 2*5^1-x =7 ?
15. (Original post by mariejuana)
What about 5^x + 2*5^1-x =7 ?
You can easily solve that and find an exact solution.
16. (Original post by RDKGames)
You can easily solve that and find an exact solution.
how ? xD
17. (Original post by mariejuana)
how ? xD
Multiply through by to get which is a quadratic.
18. (Original post by RDKGames)
Multiply through by to get which is a quadratic.

i don't understand it at allllllllllllll
19. (Original post by mariejuana)
i don't understand it at allllllllllllll
Ok, it's literally nothing more difficult than solving a quadratic. Let then you have which is the same as which I'm sure you can solve before convering back to
20. (Original post by mariejuana)
i don't understand it at allllllllllllll
i do. doesn#t matter i was beingstupud

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 18, 2017
Today on TSR

### Results day under a month away

How are you feeling?

Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE