paulo1913 Posted September 13, 2008 Share Posted September 13, 2008 how would I solve this? 13^4x-5 = 6 because I thought that 13 and 6 would have had to have a common power... Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted September 13, 2008 Share Posted September 13, 2008 Do you mean [math]13^{4x} - 5 = 6[/math] or [math]13^{4x - 5} = 6[/math]? Either way, actually, it's just a logarithm. To get the 4x "down" you need to take the logarithm of both sides. Remember that [math]\log_3 (3^x) = x[/math] and so on. Link to comment Share on other sites More sharing options...
paulo1913 Posted September 13, 2008 Author Share Posted September 13, 2008 woops i mean 13^(4x-5)=6 so would i go log(4x-5)13=log6? Link to comment Share on other sites More sharing options...
ydoaPs Posted September 13, 2008 Share Posted September 13, 2008 No. You'll also need the change of base formula if your calculator doesn't let you choose your own base. Link to comment Share on other sites More sharing options...
paulo1913 Posted September 13, 2008 Author Share Posted September 13, 2008 I don't quite understand... Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted September 13, 2008 Share Posted September 13, 2008 You'd have to take the base 13 logarithm of both sides. [math]\log_{13}(13^{4x-5}) = 4x -5[/math] If you calculator can't do base 13 logs (to find [math]\log_{13}(6)[/math]), you can do this: [math]\log_n(a) = \frac{\log_{10}(a)}{\log_{10}(n)}[/math] Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now