Basically, I don't get exponential equations. Could someone please help me? These are the two I'm having the most trouble with:

2^(1 - 2x) = 3x

(7/9)^x=11^1-x

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- Nov 21st 2011, 05:30 AMhotpotato1092Exponential functions help
Basically, I don't get exponential equations. Could someone please help me? These are the two I'm having the most trouble with:

2^(1 - 2x) = 3x

(7/9)^x=11^1-x - Nov 21st 2011, 06:44 AMSironRe: Exponential functions help
To solve the second equation you can write:

$\displaystyle \left(\frac{7}{9}\right)^x=11\cdot 11^{-x}$

$\displaystyle \Leftrightarrow \left(\frac{7}{9}\right)^x=\frac{11}{11^{x}}$

$\displaystyle \Leftrightarrow \left(\frac{7}{9}\right)^x \cdot 11^x=11$

(You can also take the logarithm of both side in the beginning and continue)

Now use (the general rule) $\displaystyle a^{x}\cdot b^{x}=(a\cdot b)^x$

About the first equation, are you sure it isn't $\displaystyle 2^{1-2x}=3^x$ in stead of $\displaystyle 2^{1-2x}=3x$? - Nov 21st 2011, 08:04 AMHallsofIvyRe: Exponential functions help
Do you understand that the

**logarithm**is the inverse function to the exponential? And, it turns out you can use logarithm to any base!

You also need to learn the "rules of expentials":

a^x a^y= a^(x+ y) and (a^x)^y= a^(xy)

and the corresponding "rules of exponentials:

log(ab)= log(a)+ log(b) and log(a^b)= b log(a).

It is that last one that is of the most use.

Taking the logarithm of both sides of that second equation, as Siron suggests, gives

log((7/9)^x)= x log(7/9)= (1-x)log(11). You now have a linear equation to solve for x.

I agree with Siron also that you should look again at the first problem. Although that**can**be solve, I doubt it is what is intended.

2^(1- 2x)= 2(2^(-2x))= 2(1/4)^x= 3x. dividing both sides by 3 and Multiplying both sides by (1/4)^x you get x4^x= 2/3. Of course, 4= e^ln(4) so 4^x= e^(xln(4)) and the equation is xe^(xln(4))= 2/3. Now multiply both sides by ln(4) so the equation is (xln(4))e^(xln(4))= 2ln(4)/3. Almost there! Let y= x ln(4). The equation ye^y= 2ln(4)/3 has solution y= W(2ln(4)/3) where "W(x)" is the "Lambert W function"- which is defined as the inverse function to f(x)= xe^x. Since y= W(2ln(4)/3) and y= x ln(4), x= W(2ln4)/3)/ln(4).

But that is NOT the sort of problem anyone who is not in a "special functions" course where they are introduced to the W function would be expected to do!

If the problem is in fact 2^(2x-1)= 3^x, take the logarithm of both sides:

(1- 2x)log(2)= x log(3). Again, you have a linear equation to solve for x.