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

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Answer 1 · 2016-04-05T11:27:58

#x=1/2#

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#