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How do you solve #2^x= 4^(x+1)#?

Answer 1

Ariana Alvarez

#color(blue)(x=-2#

#2^x=4^(x+1)# We know that #4=2^2#

So the expression can be written as: #2^x=2^(2*(x+1))#

#2^x=2^(2x+2)#

Now as the bases are equal we equate the exponents to find #x#

#x=2x+2#

#-2=2x-x#

#color(blue)(x=-2#

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Answer 2

Liam Brennan

To solve the equation (2^x = 4^{x+1}), you can use the properties of exponents to rewrite (4) as (2^2). Then, solve for (x). Here's the solution:

[2^x = (2^2)^{x+1}]

[2^x = 2^{2(x+1)}]

Since the bases are the same, the exponents must be equal:

[x = 2(x+1)]

Now, solve for (x):

[x = 2x + 2]

[x - 2x = 2]

[-x = 2]

[x = -2]

So, the solution to the equation is (x = -2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.