Solve 9^(x+1)=?27?^(x-1)

Okay, the first thing to do is recall properties of exponents and how to rewrite number as a single base. What this means is if you were to rewrite 36, you could rewrite it as: 6

^{2}.Additionally if you have 81

^{x}. Then you could rewrite as either a base of 9 or a base of 3.recall that 81=9

^{2}so 81^{x}=(9^{2})^{x}: from here properties of exponents tell you to multiply the exponents together, so you get: 81^{x}=9^{2x}If you follow the same process further you will have: 3

^{4x}.With this you may look at:

9

^{x+1}=27^{x-1}and you may see that 9 and 27 can both be rewritten as the same base. Where:9=3

^{2}and 27=3^{3}Making the substitutions:

(3

^{2})^{x+1}=(3^{3})^{x-1}Remember that the exponents multiply. But more importantly, since you have the same base, you may set the exponents equal to one another:

3

^{2(x+1)}=3^{3(x-1)}can only be true if the exponents are the same:so,

2(x+1)=3(x-1)

Distributing and solving:

2x+2=3x-3

-x=-5

x=5

You may check your answer to see if this is correct and you find that it is the solution to the exponential.