SOLUTION: Hi, could you help me to solve these equations? 2^(2x+1) - 10.2^x=-12 log(4,[log(4,(log4,x))])=0 1/3.log(20,2x-1)=log(20,11)-2.log(20,(2x-1)^(1/3))+log(5,(log7,7))

Question 202660: Hi, could you help me to solve these equations?
2^(2x+1) - 10.2^x=-12
log(4,[log(4,(log4,x))])=0
1/3.log(20,2x-1)=log(20,11)-2.log(20,(2x-1)^(1/3))+log(5,(log7,7))

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'll do the first one to give you an idea on how to tackle these problems.

Start with the given equation.

Break up the exponent using the identity

Raise 2 to the first power to get 2

Rearrange the terms.

Rewrite as using the identity

Now let . So

Replace with . Replace with

Add 12 to both sides.

Notice that the quadratic is in the form of where , , and

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

Start with the quadratic formula

Plug in , , and

Negate to get .

Square to get .

Multiply to get

Subtract from to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So the solutions (in terms of 'z') are or

Now recall that we let . So let's use this to find the solutions in terms of "x".

Start with the first solution in terms of 'z'

Plug in

Take the log of both sides

Rewrite the first log using the identity

Divide both sides by to isolate "x"

Use the change of base formula to simplify

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

Start with the second solution in terms of 'z'

Plug in

Take the log of both sides

Rewrite the first log using the identity

Divide both sides by to isolate "x"

Use the change of base formula to simplify

Evaluate the log base 2 of 2 to get 1

=======================================================================

Answer:

So the solutions are or

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