What’s the correct way to read 43.106

Mathematics

1 answer:

BabaBlast [244]4 years ago

6 0

Answer:

The correct way to read 43.106 would be forty-three and one hundred 6 hundredths.

Step-by-step explanation:

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2logbase4(x)-logbase4(5)=125

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$D:x>0\\\\ 2\log_4x-\log_45=125\\ \log_4x^2-\log_45=\log_44^{125}\\ \log_4\dfrac{x^2}{5}=\log_44^{125}\\ \dfrac{x^2}{5}=4^{125}\\ x^2=5\cdot4^{125}=5\cdot2^{250}=5\cdot(2^{125})^2\\ x=\sqrt{5\cdot2^{250}} \vee x=-\sqrt{5\cdot2^{250}}\\ x=\sqrt{5\cdot4^{125}} \vee x=-\sqrt{5\cdot4^{125}}\\ x=\sqrt{5\cdot(2^{125})^2} \vee x=-\sqrt{5\cdot(2^{125})^2}\\ x=\sqrt5 \cdot\sqrt{(2^{125})^2} \vee x=-\sqrt5 \cdot\sqrt{(2^{125})^2} \\ x=2^{125}\sqrt5 \vee x=-2^{125}\sqrt5$

$x=-2^{125}\sqrt5 \not \in D\Rightarrow \boxed{x=2^{125}\sqrt5}$

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