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I have no idea what to do at all, the book gives no relevant examples so I had to give up.

I am suppose to find $\sinh(\ln 2)$ I don't even know where to start. I think I need to use the form $\dfrac{e^{\ln2}-e^{-\ln2}}2$ but I get confused on how to do that as well.

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    Just use the fact that $e^{\ln x}=x$ for all $x>0$ and $e^{-x}=\frac 1{e^x}$ for all $x\in\mathbb R$.2011-10-06
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    Recall that logarithm is the inverse of exponentiation, i.e. for $\mathrm{e}^x = 2$, the real solution is $x= \ln 2$, now substitute the solution back into the equation (_lightbulb_)2011-10-06
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    I know I should never ask why, but is there an easy way to remember that?2011-10-06
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    "The logarithm is the inverse of the exponential" is often the *definition* of the logarithm.2011-10-06
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    So e raised to the ln2 just means e to the $loge^2$2011-10-06
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    @Jordan: No. $\ln 2 = r$ means that $e^r=2$. So $e^{\ln 2} = e^r =$`?`.2011-10-06
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    I don't know what that means.2011-10-06
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    @Jordan: $\ln$ is a function that, given an input, will return to you the exponent to which you need to raise $e$ in order to get the input. That is, if we write $\ln (3) = r$, that means that $r$ is the number such that $e^r$ is equal to $3$. If we write $\ln(5)=s$, that means that $s$ is the number such that $e^s = 5$. $\ln(e^2) = 2$, because $2$ is precisely the exponent we need to raise $e$ to in order to get $e^2$ as the answer. So when we write "$\ln 2 = r$" that means that if we do $e^r$< we will get $2$ as the result. So, if $\ln 2 = r$, then $e^{\ln 2} = e^r$. What do you get?2011-10-06
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    I don't know. ln2 I guess but I don't know why.2011-10-06
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    Jordan, see the first comment and ask yourself what '$x$' is. An 'inverse function' is something that *undoes* some original function. So if $y=e^x$, the natural logarithm 'undoes' the exponential function, i.e. $\ln y = x$.2011-10-06
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    ln is just loge?2011-10-06
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    I got the answer with my calculator I guess that is good enough.2011-10-06
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    $\ln(x)$ is $\log_e(x)$ and it's the inverse of $e^x$. You also need to know that $f(f^{-1}(x))=f^{-1}(f(x))=x.$2011-10-06
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    So logex = x? And the function of the derivative2011-10-06
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    That should be the function of the derivative is equal to the derivative of the function which is equal to x?2011-10-06

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As I've mentioned before, trying to pass calculus through sheer memorization effort is doom to fail.

Here you are asking of a way to memorize the fact that the $$e^{\ln 2} = 2.$$ This is simply a consequence of what the logarithm function is supposed to be.

Logarithms are the inverses of exponentials. Inverse functions are functions that "undo" what the "original" function does. The logarithm base $a$ "undoes" exponentiation base $a$; exponentiation base $a$ "undoes" the logarithm base $a$. We define $\log_a(b)$ to be the number $r$ exactly when $a^r = b$. The statement "$\log_a(b)=r$" and the statement "$a^r = b$" mean the same thing, just like the statement "$\frac{6}{2}=3$" and the statement "$3\times 2= 6$" mean the same thing (because "$\frac{6}{2}$" means "the number that multiplied by $2$ gives $6$").

That is, we always have $$ a^{\log_a(x)} = x\quad\text{and}\quad \log_a(a^x) = x\quad\text{for all values of }x$$

These identities are simply a consequence of the fact that each function "undoes" the other function. It is a waste of mental effort to try to memorize them: there's just too many of them! One for every possible value of $a$. It's far better (and more productive in the long run) to understand why they are true, so that they can be internalized as conclusions rather than memorized "magic spells" that have no meaning.

In particular, since $\ln x$ is the logarithm base $e$ of $x$, $\ln x = \log_e(x)$, the fact that the exponential undoes the logarithm and the logarithm undoes the exponential means that $$e^{\ln(x)} = x\quad\text{and}\quad \ln(e^x) = x\quad\text{for all values of }x.$$

So $e^{\ln 5} = 5$, $\ln(e^3) = 3$, $e^{\ln\pi} = \pi$, $17^{\log_{17}(42.4)} = 42.4$, and so on and so forth.

Since we also have that, among the properties of logarithms, $r\ln(b) = \ln(b^r)$,

So: $$\sinh(\ln(2)) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{e^{\ln 2} - e^{\ln(2^{-1})}}{2} = \frac{2 - 2^{-1}}{2} = \frac{\quad 2 - \frac{1}{2}\quad}{2} = \frac{2}{2} - \frac{\;\frac{1}{2}\;}{2} = 1 - \frac{1}{4} = \frac{3}{4}.$$

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    I still don't get $a^{\log_a(x)} = x\quad\text{and}\quad \log_a(a^x) = x\quad\text{for all values of }x$ I know I am stupid but I just can't get it for some reason. If I have 2 raised to a number that is suppose to make a number x than how do I know that equals x? That doesn't make sense to me, to me it could be anything.2011-10-06
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    The reason is that $\log_a(x)$ is **defined** to be **the** number to which you need to raise $a$ in order to get $x$ as the answer. You are raising $a$ to the number which is supposed to give you $x$ as the answer; how do you know that $x$ is the answer? Because that's what the number you are raising $a$ to is *supposed* to do.2011-10-06
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    So I need a to equal x by using an exponent. so $a^x$=x? I don't get that. If I replace a with 3 and x with 9 I get x don't I? So shouldn't it be equal to a?2011-10-06
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    $\log_a(x)$ is the only thing you can put in the empty box in $$a^{\Box} = x$$to make that equation true. There is one, and only one number that can fill that box and make the equation true. That number is what $\log_a(x)$ is defined to be equal to.2011-10-06
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    So I need to make a equal to x by raising it by an exponent, I don't understand how raising it by x makes a x. Like if I am trying to make 4 = 16 I raise it by 2 not 16.2011-10-06
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    @Jordan: Nobody said "raise it by $x$". We said: the number you need to raise $a$ to in order to get $x$ is $\log_a(x)$. That's not "x", it's **the logarithm base $a$ of $x$**. The fact that $2^{4}=16$ tells you that $\log_2(16) = 4$.2011-10-06
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    I think I get it now, hopefully I can still remember on monday. Thank you.2011-10-06