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Travka [436]
3 years ago
6

Solve 12^x-2=20 round to the nearest ten-thousandth? help, please!

Mathematics
2 answers:
luda_lava [24]3 years ago
7 0

Answer:

The answer to the equation: 12^x-2=20 round to the nearest ten-thousandth is:

x=1.2440

Step-by-step explanation:

We need to find x of the equation (1)

12^x-2=20

Now, the number two goes to the other side of the equation to add

12^x=20+2\\12^x=22

We need to multiply by logarithm on both sides of the equation

Log(12^x)=Log(22)

Using the properties of the logarithms, we know:

x*Log(12)=Log(22)\\x=\frac{Log22}{Log12}

x=1.2439

The nearest ten-thousandth is:

x=1.2440

babymother [125]3 years ago
3 0
12^x - 2 = 20
12^x = 22
x log 12 = log 22 
x = (log 22)/(log 12) 

The rest is a calculator exercise. Use any base log you like, as long as both are the same base. (The right side is really the base-12 logarithm of 22, but nobody has a base-12-log key on their calculator.)
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brainly.com/question/15775372

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