How do you solve 4^(x+1)= 4^x+6?
1 Answer
Apr 5, 2016
Explanation:
Using the rule:
a^(b+c)=a^b(a^c)
We can rewrite
4^(x+1)=4^x(4^1)=4(4^x)
This makes the equation
=>4(4^x)=4^x+6
Subtract
=>4(4^x)-4^x=6
Note that although this looks different than you may be used to,
=>3(4^x)=6
Divide both sides by
=>4^x=2
Write
=>(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