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Aleonysh [2.5K]
3 years ago
12

Can someone give me a hand?

ula1" title="{5}^{2x + 3} - 84 = 0" alt="{5}^{2x + 3} - 84 = 0" align="absmiddle" class="latex-formula">
Thanks
Mathematics
1 answer:
weqwewe [10]3 years ago
8 0

First, add 84 to both sides, to isolate the exponential function.

Then, take the logarithm of both sides:

(2x+3)*log 5 = log 84

Solving for 2x+3, we get log 84 - log 5, and so

2x = -3 + log 84 - log 5.  Dividing both sides by -3, we obtain the final answer:

      -3 + log 84 - log 5

x = -----------------------------

                    2

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Sarah is carrying out a series of experiments which involve using increasing amounts of a chemical. In the  first experiment she uses 6g of the chemical and in the second experiment she uses 7.8 g of the chemical

(i)Given that the amounts of the chemical used form an arithmetic progression find the total amount of  chemical used in the first 30 experiments

(ii)Instead it is given that the amounts of the chemical used for a geometric progression. Sarah has a  total of 1800 g of the chemical available. Show that the greatest number of experiments possible satisfies the inequality: 1.3^N \leq 91 and use logarithms to calculate the value of N.

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(a)963 grams

(b)N=17

Step-by-step explanation:

(a)

In the first experiment, Sarah uses 6g of the chemical

In the second experiment, Sarah uses 7.8g of the chemical

If this forms an arithmetic progression:

First term, a =6g

Common difference. d= 7.8 -6 =1.8 g

Therefore:

Total Amount of  chemical used in the first 30 experiments

S_n=\dfrac{n}{2}[2a+(n-1)d] \\S_{30}=\dfrac{30}{2}[2*6+(30-1)1.8] \\=15[12+29*1.8]\\=15[12+52.2]\\=15*64.2\\=963$ grams

Sarah uses 963 grams in the first 30 experiments.

(b) If the increase is geometric

First Term, a=6g

Common ratio, r =7.8/6 =1.3

Sarah has a total of 1800 g

Therefore:

Sum of a geometric sequence

S_n=\dfrac{a(r^N-1)}{r-1} \\1800=\dfrac{6(1.3^N-1)}{1.3-1} \\1800=\dfrac{6(1.3^N-1)}{0.3}\\$Cross multiply\\1800*0.3=6(1.3^N-1)\\6(1.3^N-1)=540\\1.3^N-1=540\div 6\\1.3^N-1=90\\1.3^N=90+1\\1.3^N=91

Therefore, the greatest possible number of experiments satisfies the inequality

1.3^N \leq 91

Next, we solve for N

Changing 1.3^N \leq 91 to logarithm form, we obtain:

N \leq log_{1.3}91\\N \leq \dfrac{log 91}{log 1.3}\\ N \leq  17.19

Therefore, the number of possible experiments, N=17

3 0
3 years ago
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