e^{4x+1}= 2e^{x}
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Originally Posted by Scoplex e^{4x+1}= 2e^{x} Can you solve
x = [log2 - 1]/3 ??? Does this mean that log(2) is equal to 2e? I would then do the following: 4x+1=x 3x=-1 x=-1/3 ??? Is this then correct? This logs and stuff get my head cooking sometimes. Thanks in advance.
Originally Posted by Scoplex x = [log2 - 1]/3 ???CORRECT Does this mean that log(2) is equal to 2e? What do you mean by that? The is just , no more, no less.
You already have the answer as x = [log(2) - 1]/3.
Hello, Scoplex! Take logs: . . . . . . . . . . . . . . . . Note that n . . . . . . . . . n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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