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zaharov [31]
3 years ago
7

Log base 32 of 2 Help

Mathematics
1 answer:
blagie [28]3 years ago
4 0

Answer:

log_{32}(2)=0.2

Step-by-step explanation:

Recall that the unknown (x) here is the log base 32 of 2, so we can write this as the equation:

log_{32}(2)=x

The above equation can be solved by the "change of base formula":

x=\frac{log(2)}{log(32)} \\x=0.2

We can also answer this by trying to solve the exponential equation:

32^x=2

Where we are asked what is the exponent (x) at which we need to raise the base (32) in order to obtain the answer "2"?

Notice that since

2^5=32

Then 32^{1/5} = 2

And 1/5 = 0.2 which also agrees with our previous answer

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monitta
15.        3x - 2y = -2
      3x - 3x - 2y = -3x - 2
                   -2y = -3x - 2
                    -2         -2
                       y = 1.5x + 1

y - y₁ = m(x - x₁)
 y - 3 = ⁻²/₃(x - (-2))
 y - 3 = ⁻²/₃(x + 2)
 y - 3 = ²/₃(x) - ²/₃(2)
 y - 3 = ⁻²/₃x - 1¹/₃
   + 3            + 3
       y = ⁻²/₃x + 1²/₃

16. 230 = 0.2s + 150
    - 150             - 150
        80 = 0.2s
        0.2    0.2
      400 = s

17. y = 2x + 2
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