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ladessa [460]
3 years ago
14

Match the y-coordinate with the given x-coordinate for the equation y = log10x.

Mathematics
2 answers:
mina [271]3 years ago
7 0
1. log {10 (1/100) } = log {1/10} = -1

2. log {10 (1/10)} = lg {1} = 0

3. log {10 (1)} = lg (10) = 1

4. log {10 (10)} = log {100} = 2

5. log {10 (100)} = log{1000} = 3
Andrew [12]3 years ago
5 0

The non-algebraic functions are called transcendental functions. This include the logarithmic function. The definition of Logarithmic Function with Base a is as follows:


For \ x\ \textgreater \ 0, \ a \ \textgreater \ 0, \ and \ a \neq 1 \\ \\ y=log_{a}x \ if \ and \ only \ if \ x=a^{y} \\ \\ Then: \\ \\ f(x)=log_{a}x \\ \\ is \ called \ the \ logarithmic \ function \ with \ base \ a.


We know that the equations is:


y=log(10x)


So let's solve each case:


Case 1


x=\frac{1}{100} \\ \\ y=log(10(\frac{1}{100})) \\ \\ \therefore y=log(\frac{1}{10}) \\ \\ \therefore y=-1 \\ \\ So: \\ \\ \boxed{\ x=\frac{1}{100}} \ matches \ to \ \boxed{y=-1}


Case 2


x=\frac{1}{10} \\ \\ y=log(10(\frac{1}{10})) \\ \\ \therefore y=log(1) \\ \\ \therefore y=0 \\ \\ So: \\ \\ \boxed{\ x=\frac{1}{10}} \ matches \ to \ \boxed{y=0}


Case 3


x=1 \\ \\ y=log(10(1)) \\ \\ \therefore y=log(10) \\ \\ \therefore y=1 \\ \\ So: \\ \\ \boxed{\ x=1} \ matches \ to \ \boxed{y=1}


Case 4


x=10 \\ \\ y=log(10(10)) \\ \\ \therefore y=log(100) \\ \\ \therefore y=2 \\ \\ So: \\ \\ \boxed{\ x=10} \ matches \ to \ \boxed{y=2}


Case 5


x=100 \\ \\ y=log(10(100)) \\ \\ \therefore y=log(1000) \\ \\ \therefore y=3 \\ \\ So: \\ \\ \boxed{x=100} \ matches \ to \ \boxed{y=3}

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Just find the slope between the points

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Use the chain rule calculate dw/dr , dw/ds and dw/dt
Liono4ka [1.6K]

Answer:

\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}, \frac{dw}{ds} = \frac{10\cdot s}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}, \frac{dw}{dt} = -\frac{2\cdot t}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}

Step-by-step explanation:

We proceed to derive each expression by rule of chain. Let be w = \ln (x+2\cdot y + 3\cdot z), x = r^{2}+t^{2}, y = s^{2}-t^{2} and z = r^{2}+s^{2}:

\frac{dw}{dr} = \frac{\frac{dx}{dr}+2\cdot \frac{dy}{dr} +3\cdot \frac{dz}{dr}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{dr} = 2\cdot r

\frac{dy}{dr} = 0

\frac{dz}{dr} = 2\cdot r

\frac{dw}{dr} = \frac{8\cdot r}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (1)

\frac{dw}{ds} = \frac{\frac{dx}{ds}+2\cdot \frac{dy}{ds} +3\cdot \frac{dz}{ds}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{ds} = 0

\frac{dy}{ds} = 2\cdot s

\frac{dz}{ds} = 2\cdot s

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\frac{dw}{ds} = \frac{10\cdot s}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (2)

\frac{dw}{dt} = \frac{\frac{dx}{dt}+2\cdot \frac{dy}{dt} +3\cdot \frac{dz}{dt}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{dt} = 2\cdot t

\frac{dy}{dt} = -2\cdot t

\frac{dz}{dt} = 0

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\frac{dw}{dt} = -\frac{2\cdot t}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (3)

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