M marlo03 New member Joined Sep 11, 2010 Messages 8 Sep 11, 2010 #1 Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this. log[5] (5x - 1) = 1
Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this. log[5] (5x - 1) = 1
M Mrspi Senior Member Joined Dec 17, 2005 Messages 2,116 Sep 11, 2010 #2 marlo03 said: Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this. log[5] (5x - 1) = 1 Click to expand... Here's a definition which might help you: log[sub:2n63nyhx]b[/sub:2n63nyhx] n = a MEANS this: b[sup:2n63nyhx]a[/sup:2n63nyhx] = n Now, see what that gives you when you apply it to YOUR problem: log[sub:2n63nyhx]5[/sub:2n63nyhx] (5x - 1) = 1 5[sup:2n63nyhx]1[/sup:2n63nyhx] = (5x - 1) Does that give you an equation you can solve for x? And does the value you get for x CHECK?
marlo03 said: Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this. log[5] (5x - 1) = 1 Click to expand... Here's a definition which might help you: log[sub:2n63nyhx]b[/sub:2n63nyhx] n = a MEANS this: b[sup:2n63nyhx]a[/sup:2n63nyhx] = n Now, see what that gives you when you apply it to YOUR problem: log[sub:2n63nyhx]5[/sub:2n63nyhx] (5x - 1) = 1 5[sup:2n63nyhx]1[/sup:2n63nyhx] = (5x - 1) Does that give you an equation you can solve for x? And does the value you get for x CHECK?
M marlo03 New member Joined Sep 11, 2010 Messages 8 Sep 11, 2010 #3 Yes, thank you so very very much!!!
