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4 years ago

How do you solve the function y = 2.5^x for y=20

1 answer:

6 0

$y=2.5^x;\ y=20\\\\2.5^x=20\\\\log_{2.5}2.5^x=log_{2.5}20\\\\x=log_{2.5}20\\\\x=log_{2.5}(2.5\cdot8)\\\\x=log_{2.5}2.5}+log_{2.5}8\\\\x=1+log_{2.5}8$

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What is the solution of the equation when solved over the complex numbers?

kumpel [21]

Answer:

$x=+i\sqrt{20}\approx+4.472i$

$x=-i\sqrt{20}\approx-4.472i$

Step-by-step explanation:

Starting from $x^2+20=0$ , we do $x^2=-20$ , which means $x=\pm\sqrt{-20}$ , which can be written as $x=\pm\sqrt{(-1)(20)}$ , which gives us $x=\pm\sqrt{(-1)}\sqrt{(20)}$ , which we know is $x=\pm i \sqrt{(20)}$ , so our solutions are:

$x=+i\sqrt{(20)}$

$x=-i\sqrt{(20)}$

Which can be approximated to:

$x\approx+4.472i$

$x\approx-4.472i$

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