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I am blanking on how to solve this type of equation. I have solved it using a graph, so I know the correct answer, but I need to solve it using log as well. EX: $1400e^{.046t}$ and $375e^{.352t}$.

Using a graph I got $t=4.304$, but I am unsure how to find their intersection using logarithmic functions. I think the multiplication instead of addition/subtraction is throwing me off

I started with: $$1400(0.046t)\ln e=375(.352t)\ln e$$ $$1400(0.046t)=375(.352t)$$

I tried dividing both sides, subtracting, I know there is just one thing I am missing somewhere, but every thing I try I get different answers.

Thanks for your help!

Gibbs
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Chloe Fields
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  • Use MathJax for formatting -- https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Green.H May 20 '18 at 19:45

2 Answers2

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Your result is wrong because the correct rule for the logarithm of a product is $$ \log(ab)=\log a+\log b $$ so, from $$ 1400e^{0.0467 t}=374e^{0.352 t} $$ using the rule, we have $$ \log\left(1400e^{0.0467 t}\right)=\log\left(374e^{0.352 t}\right) \iff \log 1400 +0.0467 t=\log 375 +0.352 t $$

can you complete?

Emilio Novati
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If $a = 0.046$ and $b = 0.352$ you end up with $1400e^{at} = 375e^{bt}$, which can be written as $e^{(b-a)t} = \frac{1400}{375}$. So applying the natural logarithm to both sides $$(b-a)t = \ln \frac{1400}{375} \Rightarrow t = \frac{1}{0.352-0.046}\ln \frac{1400}{375} \sim 4.305$$ which, I would say, confirms the result you get with the graph.

Gibbs
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