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

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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natagutty natagutty · Answer 2 · 2019-09-09T20:30:52-04:00

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:

\(x=1.2440\)