How do you solve #2^(2x+1)= 3(2^x)-1#?

1 Answer
John D.
Oct 13, 2017

#x = 0#

Explanation:

This looks like it could be a quadratic in terms of #color(red)(2^x)#. Let's see if we can get it that way!

With equations like this, it's best to try substituting in other variables to make the equation easier to understand. In this particular case, I see a lot of stuff relating to #color(red)(2^x)#, so let's do this:

Make a new variable #color(blue)w# which equals #color(red)(2^x)#.

Now replace #color(red)(2^x)# with #color(blue)w# wherever possible in the equation.

We can spot one #color(red)(2^x)# right off the bat:

#2^(2x+1) = 3(color(red)(2^x)) - 1#

#2^(2x+1) = 3color(blue)w - 1#

However, the other one is a little harder to spot. Let's look back at our rules of exponents:

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

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

Using these rules, we can manipulate the left side of the equation a bit:

#2^(2x+1) = 3color(blue)w - 1#

#2^1 * 2^(2x) = 3color(blue)w - 1#

#2 * (2^x)^2 = 3color(blue)w - 1#

See the other #2^x#? Now we can substitute another #w#, and then factor this equation like a quadratic!

#2 * (color(red)(2^x))^2 = 3color(blue)w - 1#

#2color(blue)w^2 = 3color(blue)w-1#

#2color(blue)w^2 - 3color(blue)w + 1 = 0#

#(2color(blue)w + 1)(color(blue)w-1) = 0#

#(2color(blue)w+1) = 0 " " or " " (color(blue)w-1)=0#

#color(blue)w = -1/2 " " or " "color(blue)w = 1#

Now, to complete the solution, we substitute #color(red)(2^x)# back in and solve it.

#color(red)(2^x)= -1/2 " " or " " color(red)(2^x) = 1#

The function #f(x) = color(red)(2^x)# is never negative, so we know that #-1/2# cannot be a solution. Therefore, we only have one solution:

#color(red)(2^x) = 1#

#x = log_2(1)#

#x = 0#

Final Answer

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