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daser333 [38]
3 years ago
15

Given that 2^x = 7, find the value of 4^x-1.

Mathematics
1 answer:
Helga [31]3 years ago
4 0

4=2^2\\\\4^x=(2^2)^x=(2^x)^2=7^2=49\\\\4^x-1=49-1=48


If\ is\ 4^{x-1},\ then\\\\4^{x-1}=\dfrac{4^x}{4}=\dfrac{49}{4}=12\frac{1}{4}

Used:\\\dfrac{a^n}{a^m}=a^{n-m}

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7 0
1 year ago
(1+1/3)^2−2/9 I am about to have a mental breakdown, please somebody answer
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Answer:

\frac{14}{9} = 1\frac{5}{9} \\\\1.5

Step-by-step explanation:

(1+1/3)^2-2/9\\\\= \left(1+\frac{1}{3}\right)^2-\frac{2}{9}\\\\\left(1+\frac{1}{3}\right)^2=\frac{4^2}{3^2}\\\\=\frac{4^2}{3^2}-\frac{2}{9}\\\\\frac{4^2}{3^2}=\frac{16}{9}\\\\=\frac{16}{9}-\frac{2}{9}\\\\\mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\\\=\frac{16-2}{9}\\\\\mathrm{Subtract\:the\:numbers:}\:16-2=14\\=\frac{14}{9}

8 0
4 years ago
Read 2 more answers
Please Help ASAP!
Mice21 [21]

Replace the range for f(x)

So...

0=4-2x

x=2

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8=4-2x

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3 years ago
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UkoKoshka [18]
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