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2 years ago

Please solve this exponential equation 3^2x-4(3^x+1)+27=0

Mathematics

1 answer:

Mumz [18]2 years ago

5 0

3²x-4(3²x+1)+27=0

9x-4(9x+1)+27=0

9x-4×9x-4×1+27=0

9x -36x -4 +27=0

9x-36x= 4-27

-27x= -23

x= -23/-27

x= 23/27

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3 years ago

Read 2 more answers

If a ladder is 5.5 m long and it is placed 3 m from the wall, solve and show the work when you use the Pythagorean Theorem to fi

natali 33 [55]

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Step-by-step explanation:

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2 years ago

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2 years ago

Express p in terms of q if

Irina-Kira [14]

$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$

8 0

2 years ago