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alekssr [168]
4 years ago
9

What’s the answer to the top question and steps how to solve?

Mathematics
1 answer:
Leviafan [203]4 years ago
3 0

Answer:

-3.69

Step-by-step explanation:

Exponential equations can be solved by taking Ln (natural logarithm) of both sides. The rule we need to know is:

Ln(a^b)=bLn(a)

Now lets work out this problem:

16^{2x+3}=13^{x-1}\\Ln(16^{2x+3})=Ln(13^{x-1})\\(2x+3)Ln(16)=(x-1)Ln(13)

Using calculator, we take the values of Ln 16 and Ln 13 as:

Ln 16 = 2.7726

Ln 13 = 2.5649

Now, we can solve for x:

(2x+3)Ln(16)=(x-1)Ln(13)\\(2x+3)2.7726=(x-1)2.5649\\5.5452x+8.3178=2.5649x-2.5649\\2.9803x=-10.8827\\x=-3.69

x is -3.69 (rounded to 2 decimal places)

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Determine the zeros of the function <br> 5) y=-x² + 2x + 1
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<em><u>Solution:</u></em>

<em><u>We have to find the zeros of the function</u></em>

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\mathrm{For\:}\quad a=-1,\:b=2,\:c=1\\\\x  =\frac{-2\pm \sqrt{2^2-4\left(-1\right)1}}{2\left(-1\right)}

Simplify\\\\x=\frac{-2 \pm \sqrt{4+4}}{-2}\\\\x =\frac{-2 \pm \sqrt{8}}{-2}\\\\Simplify\\\\x =\frac{-2 \pm 2 \sqrt{2}}{-2}\\\\x = 1 \pm \sqrt{2}

We have two zeros

x = 1 + \sqrt{2} \text{ and } x = 1 - \sqrt{2}

Thus zeros of function are x = 1 + \sqrt{2} \text{ and } x = 1 - \sqrt{2}

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