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

##### 1 Answer

#### Explanation:

This looks like it could be a quadratic in terms of

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

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

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

The function

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

#x = log_2(1)#

#x = 0#

*Final Answer*