1.) xe^-2x + 2e^-2 = 0 2.) x^2 ln x - 9 ln x = 0 3.) 2 ln 3x - 1 = 1 Thanks in advance for any help!
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The answer to the last one is . Add the one to the right side; divide by two, and then exponentiate each side. Do you follow that? As far as the first two are concerned... 2) Solve for x:
Last edited by colby2152; Jan 21st 2008 at 12:05 PM.
No3. No2.
Last edited by Singular; Jan 23rd 2008 at 09:38 PM.
Originally Posted by colby2152 2) Solve for x: Little mistake here Central solution should be Originally Posted by toop 3.) 2 ln 3x - 1 = 1 This is Can you take it from there? Anyway use parenthesis please, it could be (See my signature.)
Originally Posted by colby2152 The answer to the last one is . Add the one to the right side; divide by two, and then exponentiate each side. Do you follow that? As far as the first two are concerned... 2) Solve for x: Hi, only a short notice: The ln()-function is not definde for negative numbers. Therefore x = -3 is not a solution of this equation.
Oops! I posted the problem wrong for #3... It should have been 3.) 2 ln 3x = 1 Sorry about that!
I was even thinking that when I did the problem, but I still kept x = 0 as a solution. To the OP, you cannot take the natural log of a number zero or less.
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