Exponential Equation

[nycmath]

Solve 81^(x-4) = 9^(3x) for x.

Let me see....

On the left side, we have a base of 81.
On the right side, we have a base of 9.

We want the same base on both sides.
How is this done?

Rewrite 81 as 9^2.
We now have (9^2)^(x-4) = 9^(3x).

On the left side, apply the distributive rule on the exponent.
Let's do that separately.

2(x-4) = 2x - 8 = new left side exponent.
We have this:

9^(2x-8) = 9^(3x)

Notice that we now have the same base 9 on both sides. This is what we wanted. The next step is to simply DROP the exponent on both sides like this:

2x - 8 = 3x

Simply solve the linear equation for x.

2x - 3x = 8

-x = 8

Divide both sides by -1.

-x/(-1) = 8/(-1)

x = -8

The answer is -8.

Why don't you try one?

Solve 5^(x + 4) = 25^(4x) for x.

[Eigenvalue]

[tex]5^{x+4 }[/tex]=[tex]25^{4x }[/tex]

Rewrite 25 as 5² and apply exponent rules:
[tex]5^{x+4 }[/tex]=[tex]5^{2(4x) }[/tex]

Simplify: [tex]5^{x+4}[/tex]=[tex]5^{8x }[/tex]

As the 2 exponential expressions have the same bases and are equal, the exponents are equal:
x+4=8x

Solve for x:
7x=4
x=[tex]\frac{4}{7}[/tex]

[nycmath]

[tex]5^{x+4 }[/tex]=[tex]25^{4x }[/tex]

Rewrite 25 as 5² and apply exponent rules:
[tex]5^{x+4 }[/tex]=[tex]5^{2(4x) }[/tex]

Simplify: [tex]5^{x+4}[/tex]=[tex]5^{8x }[/tex]

As the 2 exponential expressions have the same bases and are equal, the exponents are equal:
x+4=8x

Solve for x:
7x=4
x=[tex]\frac{4}{7}[/tex]

Nicely-done! Beautiful. To the point! Terrific!