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# logarithm question watch

1. Solve to find x in terms of a where a>0 and a isnt 100

a^x=10^2x+1 the 2x+1 is all raised. I'm not reall too sure how to start the question i know the rules of logs just don't know how to apply it here.
2. Take the logarithm to base 10 of both sides (or any base, but base 10 simplifies it a bit). Then use the rule that and rearrange to solve for .
3. Take logs. What've you got?
4. Take logs of each side, expand the bracket on the right, and then it's just moving stuff around.
5. (Original post by nuodai)
Take the logarithm to base 10 of both sides (or any base, but base 10 simplifies it a bit). Then use the rule that and rearrange to solve for .
is this right log10a^x=log1010^2x+1

then does it become xloga=2x+1 log1 not sure if this bit is correct
6. Nasty question lol...

did you get it down to... "x/(2x+1) = 1/loga" ?
7. (Original post by hazbaz)
is this right log10a^x=log1010^2x+1

then does it become xloga=2x+1 log1 not sure if this bit is correct
Not quite, no; the RHS should be . Or, better, just use the fact that (by the definition of a logarithm).
8. (Original post by nuodai)
Not quite, no; the RHS should be . Or, better, just use the fact that (by the definition of a logarithm).
Therefore xloga = 2x+1 ?? lol im confused aswell
9. (Original post by Extricated)
Therefore xloga = 2x+1 ?? lol im confused aswell
Yup.
10. (Original post by nuodai)
Yup.
Sorry, this may sound like a stupid question (tired), but whats the next step
11. (Original post by Extricated)
Sorry, this may sound like a stupid question (tired), but whats the next step
How would you solve the equation for x (where p,q,r are constants)? This is no different.
12. (Original post by nuodai)
How would you solve the equation for x (where p,q,r are constants)? This is no different.
px = qx + r

px-qx = r

x(p-q) = r

x = r/(p-q)

so in this case :

xloga = 2x+1

xloga - 2x = 1

x(loga-2) = 1

X = 1/loga -2
13. (Original post by Extricated)
px = qx + r

px-qx = r

x(p-q) = r

x = r/(p-q)

so in this case :

xloga = 2x+1

xloga - 2x = 1

x(loga-2) = 1

x = 1/loga-2

is A in this case 10? maybe thats where im going wrong..wow i feel so dumb
I don't know how much I want to say because you're kind of hijacking the OP's thread here, but basically... you're not solving for a. The constant a is fixed; instead you're trying to find x in terms of a... which is what you just did. All we're told is that and , so a could be 10 or it could be 2 or 3 or even .
14. (Original post by nuodai)
I don't know how much I want to say because you're kind of hijacking the OP's thread here, but basically... you're not solving for a. The constant a is fixed; instead you're trying to find x in terms of a... which is what you just did. All we're told is that and , so a could be 10 or it could be 2 or 3 or even .
Cool.
15. i did it anyway :P thanks i'm just not that good n logarthmic equations going to practice them lat on

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