c1 question: find the value of y such that 4^y+1=8^2y-1?

funkybinoculars

It's a question in one of the solomon papers. I have the answer but I don't understand how it comes to y=5/4?

Reply 1

NotNotBatman

Original post by funkybinoculars

It's a question in one of the solomon papers. I have the answer but I don't understand how it comes to y=5/4?

Change it, so they're in the same base.

Reply 2

TheALevelStudent

Original post by funkybinoculars

It's a question in one of the solomon papers. I have the answer but I don't understand how it comes to y=5/4?

log everything

log(4)^y+1 + log(1) = log(8)^2y-1

use log rules

y+1log(4) + log(1) = 2y-1log(8)
log 1 is zero so:
y+1log(4) = 2y-1log(8)
divide both sides by log(4) and 2y-1

y+1/2y-1 = log(8)/log(4)
y+1/2y-1 = 3/2

multiply both sides by 2y-1 and 2
2y+2 = 6y-3
4y=5
y=5/4

Reply 3

funkybinoculars

Original post by TheALevelStudent

yeah i thought about using log but don't know if i'd get a mark because we don't learn logs in c1?

Reply 4

crashMATHS

Original post by TheALevelStudent

No need. It is C1. Simple indices will resolve the issue here.

Original post by funkybinoculars

It's a question in one of the solomon papers. I have the answer but I don't understand how it comes to y=5/4?

Consider:

2^x = 2^3

It shouldn't be difficult to see that x must be 3, because otherwise the sides won't be equal.

In the case of

4^{y+1}= 8^{2y-1}

Consider what base 4 and 8 have in common and convert both of them to that base, i.e if I had 9 I could write it as 3^2.

Once you have the same base, you know that you powers have to be the same (otherwise you won't have equality).

Can you go from here? :smile: