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3 years ago

Use the laws of logarithms and the values given below to evaluate the logarithmic expression.

Mathematics

2 answers:

salantis [7]3 years ago

8 0

Answer:

Step-by-step explanation:

We can use the logarithm property shown below to evaluate this:

$Log(x*y*z)=logx+logy+logz$

Now we can write Log 12 as Log (2*2*3).

Using the rule, we can write:

Log (2*2*3) = Log 2 + Log 2 + Log 3

We are given values of Log 2 and Log 3, plugging these we get:

Log (2*2*3) = Log 2 + Log 2 + Log 3

Log (2*2*3) = 0.3010 + 0.3010 + 0.4771 = 1.0791

Answer choice B is right.

Oliga [24]3 years ago

5 0

We can factor out 12 using 2 and 3.

log(12) = log(2^2 * 3)

Rewrite it using the product rule [ log(xy) = log(x) + log(y) ]

log(2^2 * 3) = 2 log(2) + log(3)

Simplify using the given values.

2(0.3010) + 0.4771

0.602 + 0.4771

1.0791

Therefore, the answer is b. ≈ 1.0791

Best of Luck!

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3 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

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Answer:

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Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

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We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

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Answer:

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