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I'm going through maths textbooks at a rather fast pace at the moment as I have been accepted to take my chemical engineering PHD in Melbourne next year.

I have been doing really well at the log rules and applying these, surprising myself in some of the problems I've solved.

I've come across two new problems that are polar opposites, but I'm having some trouble (I went completely off course with these, nothing like the textbook answer).

The first is:

$5e^{3x}= 0.3$

The second is:

$3\ln(2x-1) = 6$

I am to answer to $2$ decimal places and I'm just trying to find the smoothest solution. I'm still looking now.

Roby5
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New Zealand's finest
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  • You need to be more careful with parentheses. 5e^3x should be read as $5(e^3)x$ from the order of operations, but I suspect you mean $5e^{3x}$. MathJax can help. – Ross Millikan May 12 '16 at 02:34

1 Answers1

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First problem

$$ \begin{align} 5e^{3x}&=0.3\\ e^{3x}&=0.06\\ 3x&=\ln{(0.06)}\\ x&=\frac{1}{3}\ln{(0.06)}\\ &= -0.9378 \end{align} $$

Second problem

$$ \begin{align} 3\ln{(2x+1)}&=6\\ \ln{(2x+1)}&=2\\ 2x+1&=e^2\\ x&=\frac{1}{2}(e^2-1)\\ &=3.1945 \end{align} $$

andselisk
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WW1
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    For the second part, OP has $2x-1$ rather than $2x+1$, but if one just follows the steps of this solution with the obvious sign change, a correct answer is obtained ($x=4.1945$), so I gave this an upvote. – David K May 12 '16 at 02:39