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# C1/C2 Help! watch

1. Hi, im working on the following question, but i havent been taught to use log when solving this. It asks you to solve it for x. Wondered if someone could remind me how to do it without using log??

5^(2x)=1/125

Again, haven't been taught to use log, taught another way. Any help??
2. (Original post by LuckyEms)
Hi, im working on the following question, but i havent been taught to use log when solving this. It asks you to solve it for x. Wondered if someone could remind me how to do it without using log??

5^(2x)=1/125

Again, haven't been taught to use log, taught another way. Any help??
logs aren't needed here, , now what power of would give you ?
remember would give you
then once you found the power make it equal to 2x then you can rearrange to find what x is from there.
3. Take loagarithms of both sides, remembering that log(A^B) = B*log(A). There is an easier way, however - write 1/125 as a power of five and then equate this to 2x.
4. 5^3 = 125, so 5^-3 = 1/125, then you just need to divide -3 by 2 to give x on its own (-1.5)
logs aren't needed here, , now what power of would give you ?
remember would give you
5^-3 gives 1/125
6. (Original post by LuckyEms)
5^-3 gives 1/125
-3=2x now solve x.
-3=2x now solve x.
Ahh thank you
8. (Original post by LuckyEms)
Ahh thank you
you're welcome.
9. (Original post by JackS94)
5^3 = 125, so 5^-3 = 1/125, then you just need to divide -3 by 2 to give x on its own (-1.5)
don't post solutions, read the rules.
10. Thank you all!!

Mind someone helping again?

The next one is 9^x+1 = 3^x sqrt(27).

I started by doing 3^x x 27^1/2 to get 81^x+1/2

Am i going in the right direction??
don't post solutions, read the rules.
My most unreserved apologies. I've never used the maths forum before, you see, and I thought I'd just try and help, but clearly the answer to -3/2 is too much of a spoiler to be said out loud.
12. (Original post by JackS94)
My most unreserved apologies. I've never used the maths forum before, you see, and I thought I'd just try and help, but clearly the answer to -3/2 is too much of a spoiler to be said out loud.
could you help me now then? ive posted the bit im not sure about
13. (Original post by LuckyEms)
Thank you all!!

Mind someone helping again?

The next one is 9^x+1 = 3^x sqrt(27).

I started by doing 3^x x 27^1/2 to get 81^x+1/2

Am i going in the right direction??
is the question ?
14. Take logs of both sides then use the power rule to place x in front of the log and then just solve that for x
15. (Original post by gaffer dean)
is the question ?
No, the sqrt it seperate to the 3^x

so its 3^x (multiplied by) sqrt(27)

16. (Original post by LuckyEms)
Thank you all!!

Mind someone helping again?

The next one is 9^x+1 = 3^x sqrt(27).

I started by doing 3^x x 27^1/2 to get 81^x+1/2

Am i going in the right direction??
don't do that (the bit i crossed out) you want everything to the power of 3 to work out what x is,

(because ) now use the rule

So, you would get
now you can work out the rest from here.
17. (Original post by LuckyEms)
No, the sqrt it seperate to the 3^x

so its 3^x (multiplied by) sqrt(27)

Oh okay, well some hints then. Root 27 is the same as (3^3)^(1/2). 9 is 3^2.
18. (Original post by JackS94)
My most unreserved apologies. I've never used the maths forum before, you see, and I thought I'd just try and help, but clearly the answer to -3/2 is too much of a spoiler to be said out loud.
That's ok.
I'm sorry I didn't mean to be rude about it.
19. (Original post by LuckyEms)
No, the sqrt it seperate to the 3^x

so its 3^x (multiplied by) sqrt(27)

You want to try to get everything in terms of a common base, in this case 3.

So for example so you can rewrite the LHS as a power of 3.

Similarly, on the RHS you can write 27 as a power of 3, and therefore you can write sqrt(27) as a power of 3.

If you can rewrite your equation in the form

where a and b are expressions that involve x, then you can write a = b and solve the resulting equation to get x.
20. Ok thank you everyone who helped!! I will have a go at this and let you know how i get on! Thank you!!

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Updated: March 20, 2013
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