How do you solve 4^(x+1)= 4^x+6?

1 Answer
Apr 5, 2016

x=1/2

Explanation:

Using the rule:

a^(b+c)=a^b(a^c)

We can rewrite 4^(x+1) as

4^(x+1)=4^x(4^1)=4(4^x)

This makes the equation

=>4(4^x)=4^x+6

Subtract 4^x from both sides.

=>4(4^x)-4^x=6

Note that although this looks different than you may be used to, 4(4^x)-4^x=3(4^x) for the same reason that 4u-u=3u.

=>3(4^x)=6

Divide both sides by 3.

=>4^x=2

Write 4 as 2^2.

=>(2^2)^x=2

Simplify the left hand side using the rule:

(a^b)^c=a^(bc)

This yields

=>2^(2x)=2

We now have two exponential terms with the same base. Thus, since they're equal, we know their exponents must also be equal.

=>2^(2x)=2^1

=>2x=1

=>x=1/2