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Lynna [10]
3 years ago
11

Logarithmic and Exponential Forms of Equations In Exercise, write the logarithmic equation as an exponential equation, or vice v

ersa.
e - 4 = 0.0183 ...
Mathematics
1 answer:
marysya [2.9K]3 years ago
8 0

Answer : The logarithmic equation of given exponential equation is, \ln (0.0183)=-4

Step-by-step explanation :

As we are given the exponential equation.

e^{-4}=0.0183

Formula used :

The inverse property of logarithm for the expression \ln e^x is:

\ln e^x=x

Now we have to determine the logarithmic equation of given exponential equation.

e^{-4}=0.0183

Taking natural logarithm both sides as,

\ln e^{-4}=\ln (0.0183)

-4=\ln (0.0183)

\ln (0.0183)=-4

Thus, the logarithmic equation of given exponential equation is, \ln (0.0183)=-4

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A sequence is constructed according to the following rule: its first term is 7, and each next term is one more than the sum of t
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Answer:

5

Step-by-step explanation:

According to the described rule, we have

a_1=7\\ \\a_1^2=7^2=49\Rightarrow a_2=4+9+1=14\\ \\a_2^2=14^2=196\Rightarrow a_3=1+9+6+1=17\\ \\a_3^2=17^2=289\Rightarrow a_4=2+8+9+1=20\\ \\a_4^2=20^2=400\Rightarrow a_5=4+0+0+1=5\\ \\a_5^2=5^2=25\Rightarrow a_6=2+5+1=8\\ \\a_6^2=8^2=64\Rightarrow a_7=6+4+1=11\\ \\a_7^2=11^2=121\Rightarrow a_8=1+2+1+1=5\\ \\\text{and so on...}

We can see the pattern

a_5=a_8=a_{11}=a_{14}=...=5\\ \\a_6=a_9=a_{12}=a_{15}=...=8\\ \\a_7=a_{10}=a_{13}=a_{16}=...=11

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a_{3k-1}=5\\ \\a_{3k}=8\\ \\a_{3k+1}=11

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