# `3(2^x)+2=2^(2x+1)` Solve the equation.Please state your workings clearly.Thanks

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`3(2^x) +2= 2^(2x+1)`

First, rewrite the right-hand side using the rules of exponents:

`2^(2x+1) = 2^(2x) *2 = 2*(2^x)^2 `

Then use substitution `y = 2^x` . Note that the y can have only positive values because any exponent of 2 is positive. The equation becomes

`3*y +2 = 2y^2`

Write this equation in standard form:

`2y^2 - 3y-2 = 0`

The left-hand side now can be factored by grouping:

`2y^2-3y-2 = 2y^2 - 4y + y - 2 = 2y(y-2) +(y-2) = (y-2)(2y+1)`

`(y-2)(2y+1) = 0`

y - 2 = 0 or 2y+1 = 0

y = 2 and y = -1/2

Since `y=2^x` it has to be positive, so the negative solution is extraneous.

Substitute the value of y = 2 to find x:

`2 = 2^x`

**x = 1**