Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


$$ (1) \log y=e^{x}+4$$ $$(2) \frac{1}{2^{y}}=\frac{1}{2^{x}}+5$$

Please write full steps and if possible give an explantion. Thank You.

share|cite|improve this question
How to ask a homework question? – Gigili May 26 '12 at 4:56
@user32251: Please see… – user9413 May 26 '12 at 4:56
You already have a hint below. You should at least try to solve it yourself. – Eugene May 26 '12 at 5:11
It seems an oddly worded question. One doesn't really differentiate an equation, one differentiates a function. I would have expected something like "If $\log y=e^x+4$, find $\frac{dy}{dx}$." – André Nicolas May 26 '12 at 5:20
Editing many times by different users has changed the question! I can't figure out what was wrong with the first version. – Gigili May 26 '12 at 5:31
up vote 3 down vote accepted


There are two ways to solve the first one. If you are familiar with implicit differentiation then differentiate both sides to get $$\begin{align*} \frac{d}{dx}(\log y)&=\frac{d}{dx}(e^x+4)\\ \frac{1}{y}\frac{dy}{dx}&=e^x+0\\ \frac{dy}{dx}&=ye^x\end{align*}$$ If you are not familiar with implicit differentiation then write $\log y=e^x+4$ as $$y=e^{e^x+4}$$ Differentiate and get $$\begin{align*} &\frac{dy}{dx}=\frac{d}{dx}\left(e^{e^x+4}\right)\\ &\frac{dy}{dx}=e^x\cdot e^{e^x+4}=ye^x\end{align*}$$


In the second question we can also use implicit differentiation or explicitly solve for y. Using implicit we get $$\begin{align*} \frac{1}{2^y}&=\frac{1}{2^x}+5\\ \frac{d}{dx}\left(\frac{1}{2^y}\right)&=\frac{d}{dx}\left(\frac{1}{2^x}+5\right)\\ \frac{-\ln2}{2^y}\frac{dy}{dx}&=\frac{-\ln2}{2^x}\\ \frac{dy}{dx}&=\frac{2^y}{2^x}\\ \end{align*}$$ But since $2^y=\frac{1}{\frac{1}{2^x}+5}$ $$\frac{dy}{dx}=\frac{\frac{1}{\frac{1}{2^x}+5}}{2^x}=\frac{1}{1+5\cdot2^x}$$ The other approach is by explicitly solving for $y$ and then differentiating as follows: $$\begin{align*} \frac{1}{2^y}&=\frac{1}{2^x}+5\\ \Longrightarrow2^y&=\frac{1}{\frac{1}{2^x}+5}\\ \Longrightarrow\log_2(2^y)&=\log_2\left(\frac{1}{\frac{1}{2^x}+5}\right)\\ \Longrightarrow y&=\log_2 1-\log_2(\frac{1}{2^x}+5)\\ \Longrightarrow y&=-\frac{\ln(\frac{1}{2^x}+5)}{\ln2}\\ \frac{dy}{dx}&=\frac{-1}{\ln2}\frac{d}{dx}\left(\ln\left(\frac{1}{2^x}+5\right)\right)\\ \frac{dy}{dx}&=\frac{-1}{\ln2}\cdot\frac{1}{\frac{1}{2^x}+5}\cdot \frac{-\ln2}{2^x}\\ \frac{dy}{dx}&=\frac{1}{1+5\cdot2^x} \end{align*}$$

*Remember that $\displaystyle\log_b a= \frac{\log_c a}{\log_c b}$

share|cite|improve this answer
You Saw wrong second question. – Neer May 26 '12 at 5:40
@Neer I fixed it now – E.O. May 26 '12 at 6:10
@Neer This is a rather poor comment to someone who wrote you a very long and detailed answer to your badly-written question. – Phira May 26 '12 at 11:43
Sorry and thank you @E.O. – Neer Apr 6 '13 at 9:17

$$\log y= e^{x} +4$$ $$ e^{\log y}= e^{(e^{x}+4)}$$ $$\Rightarrow y= e^{(e^{x}+4)}$$ Now differntiate with respect to $x$, we get, $$ \frac{dy}{dx}= e^{(e^{x}+4)}.e^{x}$$

$$\frac{1}{2^{y}}=\frac{1}{2^{x}+5}$$ $$ \therefore 2^{y}= 2^{x}+5$$ $$ \log{2^{y}}=\log{(2^{x}+5)}$$ $$\Rightarrow y\log 2= \log{(2^{x}+5)}$$ Now differentiate with respect to $x$, we get, $$\log 2\frac{dy}{dx}= \frac{1}{2^{x}+5}.2^{x}\log 2$$ $$\therefore \frac{dy}{dx}=\frac{2^{x}}{2^{x}+5}$$

share|cite|improve this answer
Second question has changed. – Neer May 26 '12 at 5:23

For the first one, you need to exponentiation both sides of the equation to get $y = e^{(4+e^x)}$ (assuming you are finding $dy/dx$), then go to the following for an exact step-by-step solution:

For the second one, the first step is to re-arrange it as: $$\frac{1}{ 5+\frac{1}{2^x}} = 2^y$$ Then take the base to logarithm of each side to get:

$$y = \log_2 \frac{1}{ 5+\frac{1}{2^x}}$$

then to compute the derivative go to the following:


Wolfram Alpha is a great tool for little pesky problems like this.... but don't overdo it!

share|cite|improve this answer
If you use the hyperlink tool above the edit box, it usually takes care of the escaping correctly so that your link does what it should. – Dylan Moreland May 26 '12 at 5:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.