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4 years ago

13

How do you write 114,000.000.000+70,000+8,000+7 in standard form

2 answers:

In-s [12.5K]4 years ago

8 0

<span>the standard form is:
114,000,078,007</span>

uranmaximum [27]4 years ago

3 0

In the first number, I think you mean commas instead of periods.
I hope that's right.

Let me see if I can line up all the pieces correctly on the screen:

114,000,000,000
70,000
8,000
7
-----------------------------

Now, by carefully implementing the mathematical operation called "addummup",
you arrive at the conclusion, which is just the sum of the pieces:

<u>114,000,078,007 </u>.

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$2*log(x)+log(y)$

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So, there are two logarithmic identities you're going to need to know.

<em>Logarithm of a power</em>:

$log_ba^c=c*log_ba$

So to provide a quick proof and intuition as to why this works, let's consider the following logarithm: $log_ba=x\implies b^x=a$

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We get the equation: $b^{xc}=a^c$

If we convert this back into logarithmic form we get: $log_ba^c=x*c$

Since x was the basic logarithm we started with, we substitute it back in, to get the equation: $log_ba^c=c*log_ba$

Now the second logarithmic property you need to know is

<em>The Logarithm of a Product</em>:

$log_b{ac}=log_ba+log_bc$

Now for a quick proof, let's just say: $x=log_ba\text{ and }y=log_bc$

Now rewriting them both in exponential form, we get the equations:

$b^x=a\\b^y=c$

We can multiply a * c, and since b^x = a, and b^y = c, we can substitute that in for a * c, to get the following equation:

$b^x*b^y=a*c$

Using the exponential identity: $x^{a}*x^b=x^{a+b}$ , we can rewrite the equation as:

$b^{x+y}=ac$

taking the logarithm of both sides, we get:

$log_bac=x+y$

Since x and y are just the logarithms we started with, we can substitute them back in to get: $log_bac=log_ba+log_bc$

Now let's use these identities to rewrite the equation you gave

$log(x^2y)$

As you can see, this is a log of products, so we can separate it into two logarithms (with the same base)

$log(x^2)+log(y)$

Now using the logarithm of a power to rewrite the log(x^2) we get:

$2*log(x)+log(y)$

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