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# exponents and logarithms

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If x is a real number such that 2^{2x+3}=14, find $$2^x$$.

Apr 19, 2018

#1
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2^{2x+3}=14     Take log of both sides

(2x+3) Log 2 = log 14     solve for 'x'

x = (log 14/log2 -3  )/2   =  .403677461        Then 2^x = 1.322876

Apr 19, 2018
#2
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Guest Apr 19, 2018
#3
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2(2x+3)=14

The first thing you do is take a log base 2 of both sides, as exponentials and logarithms are inverses. I got the base of 2 from what the 2x+3 is raised to, which is a 2.

$$log_2(2^{2x+3})=log_2(14)$$

Since the base of the logarithm and the base of the exponential are the same, they cancel, leaving only 2x+3.

$$2x+3=log_2(14)$$

Subtract 3 from both sides.

$$2x=log_2(14)-3$$

Divide by 2

$$x=\frac{log_2(14)-3}{2}$$

$$x=0.403677461$$

So, 2x is equivalent to $$2^{0.403677461}$$.

$$2^{0.403677461}=1.322876$$

So, ElectricPavlov is right about the final answer, but I wanted to reiterate the steps in a little more detail, just for clarity's sake.

Apr 19, 2018
#4
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Thanks, EP  and AT   !!!!!!   CPhill  Apr 19, 2018
#5
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Since the question asks for 2x, not x, there is no need to take logs:

2(2x+3) = 14

22x.23 = 14

22x = 14/8   (since 23 = 8)

(2x)2 = 14/8

2x = (14/8)1/2

2x ≈ 1.322876

Apr 19, 2018