What is the value of k in the product of powers below? welp

Mathematics

2 answers:

8090 [49]3 years ago

7 0

I guess : k = -1 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

defon3 years ago

6 0

Answer: The required value of k is -1.

Step-by-step explanation: We are given to find the value of k in the product of the following powers :

$10^{-3}.10.10^k=10^{-3}~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the following properties of exponents to solve the given equation.

$(i)~a^b.a^c=a^{b+c},\\\\(ii)~a^x=a^y~~~~\Rightarrow x=y.$

From equation (i) we have

$10^{-3}.10.10^k=10^{-3}\\\\\Rightarrow 10^{-3+1+k}=10^{-3}\\\\\Rightarrow 10^{-2+k}=10^{-3}\\\\\Rightarrow -2+k=-3\\\\\Rightarrow k=-3+2\\\\\Rightarrow k=-1$

Therefore, the required value of k is -1.

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Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i

podryga [215]

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

$A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},$

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.$