Is A*B^(x+1) = A*(x+1)*lnB ?

if not how could I find x?

2. $\displaystyle AB^{(x+1)} = A(x+1)lnB$

divide both sides by A

$\displaystyle B^{(x+1)} = (x+1)lnB \Rightarrow B^{(x+1)} = lnB^{(x+1)}$

if (x+1) = 0 then x = -1

so

$\displaystyle B^{0} = lnB^{0} \Rightarrow 1 = 1$

3. consider the case where A,B and x all equal 1.

4. Originally Posted by bigwave
$\displaystyle AB^{(x+1)} = A(x+1)lnB$

divide both sides by A

$\displaystyle B^{(x+1)} = (x+1)lnB \Rightarrow B^{(x+1)} = lnB^{(x+1)}$

do you know what x has to be so these will be equal
I'm sorry, I forgot: let's say it's =2

5. ## if the exponents are 0 the term goes to 1

try

if $\displaystyle (x+1) = 0$
then $\displaystyle x = -1$

if the exponents are 0 the terms goes to 1

also check out pickslides question

6. Originally Posted by bigwave
try

if $\displaystyle (x+1) = 0$
then $\displaystyle x = -1$

if the exponents are 0 the terms goes to 1

also check out pickslides question
I know X equals 75 (more or less)
The equation I need to solve s this: