SOLUTION: Thank you so much for your help! Can you please help me solve the equation.... 42x + 4x+1 � 6 = 0 (the exponents are the 2x and x+1)

Question 143984: Thank you so much for your help! Can you please help me solve the equation....
42x + 4x+1 � 6 = 0
(the exponents are the 2x and x+1)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given equation

Rewrite as using the identity

Evaluate to get 4

Rearrange the terms

Rewrite as using the identity

Let

Plug in . In other words, replace each instance of with "u"

Let's use the quadratic formula to solve for "u":

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=4, and c=-6

Square 4 to get 16

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Now break up the fraction

or

Simplify

or

Remember, we let . So

or

Let's solve the first equation

Start with the first equation

Take the log of both sides

Rewrite the left side using the identity

Divide both sides by to isolate x

Combine the logs by use of the change of base formula

---------------------------------------------

Now let's solve the second equation

Start with the first equation

Take the log of both sides

Notice how . Since we cannot take the log of a negative number, this means we must ignore it.

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Answer:
So the solution is

which approximates to

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