keeping in mind that when the logarithm base is omitted, the base 10 is assumed.
![\textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \log(x)=2\implies \log_{10}(x)=2\implies 10^2=x\implies 100=x](https://tex.z-dn.net/?f=%5Ctextit%7Bexponential%20form%20of%20a%20logarithm%7D%20%5C%5C%5C%5C%20%5Clog_a%28b%29%3Dy%20%5Cqquad%20%5Cimplies%20%5Cqquad%20a%5Ey%3D%20b%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Clog%28x%29%3D2%5Cimplies%20%5Clog_%7B10%7D%28x%29%3D2%5Cimplies%2010%5E2%3Dx%5Cimplies%20100%3Dx)
Answer:
$10,249.80
Step-by-step explanation:
I'm really tired but if you want an explanation, please say so
Answer:
1, 4, 16, 64, 256, 1024, 4096, 16384
Step-by-step explanation:
Multiply by 4
<h3>
Answer: Yes they are equivalent</h3>
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Work Shown:
Expand out the first expression to get
(a-3)(2a^2 + 3a + 3)
a(2a^2 + 3a + 3) - 3(2a^2 + 3a + 3)
2a^3 + 3a^2 + 3a - 6a^2 - 9a - 9
2a^3 + (3a^2-6a^2) + (3a-9a) - 9
2a^3 - 3a^2 - 6a - 9
Divide every term by 2 so we can pull out a 2 through the distributive property
2a^3 - 3a^2 - 6a - 9 = 2(a^3 - 1.5a^2 - 3a - 4.5)
This shows that (a-3)(2a^2 + 3a + 3) is equivalent to 2(a^3 - 1.5a^2 - 3a - 4.5)
This one is very simple.
create a simple equation to find out
15 = 12 + x
subtract 12 from each side
x= 3
the missing number is 3
or, to be even simpler, take the total, subtract the total from the number you know and the difference is the missing number.
hope I helped
ALSO, we can check going down vertically. 3 + 10 =13. this is the total listed at the bottom.
