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**yzobel** 1) ln(2t+1)=2-lnt Solve for t.

I just want to know whether I need to move lnt to the left side in order to solve the equation or can I just leave the equation as is and multiply both sides by e.

You don't multiply by e: you APPLY e at both sides. But first move ln t to the left side:

$\displaystyle \ln((2t+1)t)=2\Longrightarrow 2t^2+t=e^2...$

2) This problem has been bothering me for a while.

Differentiate y=10^tan(theta)

I tried to solving it earlier but I got stuck:

log10y=tan(theta)

So now can I use implicit differentiation?

Indeed: $\displaystyle \frac{1}{y}dy=\frac{1}{\cos^2\theta}d\theta$ and etc.

Tonio

Thank you!