Step-by-Step Solution for Solving Exponential Equations with Fractional Bases

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Corkery
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Homework Statement


Solve the equation:
(1/9)^m = 81^m + 4


Homework Equations


I know how to do a similar equation:
9^2p = 27^p-1

3^4p = 3^3p-1

cancel out the 3's because the base is the same.

4p = 3p-1
-3p -3p

1p = -1


The Attempt at a Solution


I know how to do ones like that but when it comes to this I start to struggle with fractions. :'(.

I'm totally lost and can't even make an attempt to this equation. I'm not asking for the answer but just a maybe the first step or something to get me on the right path.
 
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How about converting some terms to powers of 9?
 
Corkery said:

Homework Statement


Solve the equation:
(1/9)^m = 81^m + 4


Homework Equations


I know how to do a similar equation:
9^2p = 27^p-1

3^4p = 3^3p-1

cancel out the 3's because the base is the same.

4p = 3p-1
-3p -3p

1p = -1


The Attempt at a Solution


I know how to do ones like that but when it comes to this I start to struggle with fractions. :'(.

I'm totally lost and can't even make an attempt to this equation. I'm not asking for the answer but just a maybe the first step or something to get me on the right path.

[tex](\frac{1}{9})^m = 81^{m+4}[/tex]

[tex](9^{-1})^m = (9^2)^{m+4}[/tex]

[tex](9)^{-m} = (9)^{2m+8}[/tex]

[tex]-m = 2m+8[/tex]

[tex]-8 = 3m[/tex]

[tex]m = - \frac{8}{3}[/tex]
 
Corkey,

The way you manage your notation is very important: you wrote
(1/9)^m = 81^m + 4


2. Homework Equations
I know how to do a similar equation:
9^2p = 27^p-1

3^4p = 3^3p-1

cancel out the 3's because the base is the same.

4p = 3p-1
-3p -3p

1p = -1
which showed your original equation of difficulty to be very different from what you meant. In simple text-based expressiveness, you needed proper grouping saying (1/9)^m = 81^(m+4); The way you first wrote it, the "4" was not included as part of the exponent.