Calculus - exponential form?

Look:

lny = x ln2

lny = ln2^x

y=2^x

The fastest way to arrive at the answer is to use the formula

log(a^b) = b.log(a) to rewrite x.ln2 as ln(2^x). The base of the log is irrelevant here, as long as you keep it the same through the transformation.

Then you will have lny = ln(2^x), which is immediately followed by y = 2^x

Now the comprehensive, more lengthy way to get the same thing:

In order to write this in exponential form, you need to isolate x to one side and a single logarithm to the other side of the equation. First, divide both sides of the original equation by ln2: (lny)/(ln2) = x, to isolate x

Now you need to know that the LHS equals log base 2 of y (*). Then since that equals x, y = 2^x

(*) The formula given in your text book is probably :( log base a of b ) x (log base b of c) = (log base a of c),

or equivalently (log base b of c) = (log base a of c)/(log base a of b).

Notice the common base in the logarithms on the right side of this expression. Also, notice that the argument in the numerator of the RHS is the argument in the LHS' logarithm, and the argument in the denominator on the RHS is the base in the LHS' logarithm.

Same for this particular example, lny is simply (log base e of y). So lny/ln2 is (log base e of y)/(log base e of 2). Following the same explanation above, this equals to (log base 2 of y).

I know this can be hard to follow, but one you know the rule firmly it will intuitively come to you. Good luck,

That's not calculus that more precalc or advanced algebra. You must know the properties of logs.

lny = x ln2

lny = ln2^x

y=2x^2

u should know these rules:

(i) lne^x = e^lnx = x in other words ln and e they always cancel each other.

(ii) ln(x^a) = a. lnx

now:

xln2 = ln2^x (using rule (ii) )

Thus lny = ln2^x, take e on both sides u'll get:

e^lny = e^(ln2^x) and remember that ln and e cancel each other:

y = 2^x

by the properties of the natural log

x ln2 = ln(2^x)

so

lny = ln(2^x)

you can cancel out the natural logs if an equation is in the form lnu=lnz. by this property

lny = ln(2^x)'

e^(lny)=2^x

e^ln(u)=u is another property so

y=2^x

OK, try this method

x ln 2 = ln (2)^x

raise both sides of this equation to the e power

y = 2^x

and that is the answer.

loga b = a^b this is the defining equation for log base a.

It is important to note that the log is an exponent and follows the rules of exponents

a^n a^m = a^(n+m)

(a^n)^m = a^(nm)

so x ln 2 that is x time ln 2 is equal to ln 2^x

I hope that helps.

∫ (5x² - 2x+1) / x² dx = ∫ (5 - 2/x + 1/x²) dx = 5x - 2ln |x| - 1/x + C I'm not sure what you mean by positive exponential form..