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# Need help with this one, Logarithms? watch

1. Hey, think I'm having a major brain freeze! Can anyone help me out with this question? Hope you can all see it OK? Thanks!

2. It's a quadratic in 8^x. Can you see why?
3. Thanks for the fast reply!

I did think it might be that actually. I can see that the first part 64^x can be written as 8^2x but I still can't do it because I don't know how to deal with the second part which is ^(x+1) and not simply ^(x)
4. (Original post by Dam)

I did think it might be that actually. I can see that the first part 64^x can be written as 8^2x but I still can't do it because I don't know how to deal with the second part which is ^(x+1) and not simply ^(x)
you must have learnt the laws of indices in core 1.

the first bit 4(64^x) can be written as 4(8^2x)
the second term can be written as 5(8 x 8^x) then multiply by 8 to get 40(8^x)

using the substitution y=8^x form a quadratic and solve.

Spoiler:
Show
4y^2 + 40y -156 = 0
y^2 + 10y - 39 = 0
(y-3)(y+13)=0

y=3 so 8^x = 3 therefore x = 0.528 (3sf)
y = -13 8^x = -13 no solution
5. I feel pretty stupid not spotting 8^(x+1) can be written as 8 * 8^(x)

I've been away from studying for a while, I got my A-level in 2003, this question is part of an access course I'm doing to get back to Uni this year, but still should have no excuses for missing that one.

Thanks for helping me out.

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Updated: April 6, 2011
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