$(\frac{1}{4})^{3z-1} = 16^z + 2*16^{z-2}\\4^{1-3z} = 4^{2z} + 4^{\frac{1}{2}}*4^{2z-4}\\4^{1-3z} = 4^{2z} + 4^{2z-4+\frac{1}{2}}\\4^{1-3z} = 4^{2z} + 4^{2z-\frac{7}{2}}\\4^{1-3z} = 4^{2z} *(1+ 4^{-\frac{7}{2}})\\4^{1-3z} = 4^{2z} *(1+ 2^{-7})\\4^{1-3z} = 4^{2z} *(1+ \frac{1}{128} )\\4^{1-3z} = 4^{2z} *(\frac{129}{128} )\\Taking\;\; Logarithm\;\; with\;\; base\;\; 4\\Log_4(4^{1-3z}) = Log_4(4^{2z}) + Log_4(\frac{129}{128})\\1-3z = 2z + 0.005613627712 \\5z = 0.994386372\\z = 0.198877274$

Hence, the value of Z = 0.198877274

8 0

3 years ago

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Determine the intercepts of the line

Natali [406]

Answer:

x-intercept (-7,0)

y-intercept (0,2)

Step-by-step explanation:

4 0

3 years ago

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At a party there are 40 cookies. There are 5 chocolate cookies for every 3 oatmeal cookies,how many oatmeal cookies are there

Pavel [41]

The answer is: 15

Because if for every 5 c cookies there is 3 oatmeal cookies, we can add 5+3=8. 40 divided by 8= 5. 5 times 3 equals 15. That’s why the answer is 15

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3 years ago

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