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Charge on a body varies with time $~t~$ as $$q=16(1-e^{-3t})+10~.$$ Find the slope of the $~q-t~$ graph at point $~(\ln 2, 24)~$

If I simply take its derivative $$\frac{dq}{dt}=16(-e^{-3t})(-3)$$ $$=48e^{-3t}$$

What should I do next, as substantiated $~t~$ doesn’t given a meaning answer.

Answer is $~6~$.

nmasanta
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Aditya
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3 Answers3

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Just substitute $t = \ln 2$ into your expression. Use the laws of logs to simplify.

$48e^{-3\ln 2} = 48(e^{\ln 2})^{-3} = 48(2^{-3}) = \frac{48}8 = 6$

Deepak
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Hint
$$e^{-3\ln a} = e^{\ln (a^{-3}) } = a^{-3}=\frac{1}{a^3}$$

AgentS
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The natural exponential function (with base $e$) and the natural logarithm function are a pair of inverse functions;

$\quad$ $\quad$ $\quad$ if $f(x) = e^x$ and $g(x) = ln(x)$, then $f(g(x)) =$ $g(f(x)) = x$.

Your question involves

$\quad$ $\quad$ $\quad 48e^{-3t} \, \bigg\rvert_{t=ln(2)} $

This is worked out as follows:

$\quad$ $\quad$ $\quad 48e^{-3ln(2)} = 48e^{(ln(2)) \cdot -3} = 48 \cdot 2^{-3} = \frac{48}{8}$

This evaluates to 6.

allan e
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