## marcelie one year ago Help please !!! Solve the following with a calculator. 2log (100^-3)

1. marcelie

|dw:1444009782868:dw|

2. anonymous

-12

3. marcelie

how you get that

4. anonymous

5. marcelie

lol z.. i wanna know know how solve it by hand.

6. Nnesha

apply the power rule

7. Nnesha

power rule $\large\rm log_b x^y = y \log_b x$

8. Nnesha

$\huge\rm 2 \log (100^{-3})$ number at front of log becomes exponent of (100^{-3})$\large\rm log (100^{-3})^2$ now multiply the exponents $\rm (X^m)^n=x^{m \times n}$ and then negative exponent rule $\large\rm x^{-m}=\frac{ 1 }{ x^m }$ then there are two ways to find final answer 1st) use change of base formula and then calculator or 2nd) set it equal to x to solve by hand

9. Nnesha

make sense ? take ur time to read that :=)

10. marcelie

one second let me attach the picture

11. marcelie

12. Nnesha

$$\huge\color{green}{\checkmark}$$

13. Nnesha

gimme a sec i'll check it again

14. Nnesha

looks right

15. marcelie

aiight got it :) do you know any website that has log formulas

16. Nnesha

quotient rule$\large\rm log_b x - \log_b y = \log_b \frac{ x }{ y}$ to condense you can change subtraction to division product rule $\large\rm log_b x + \log_b y = \log_b( x \times y )$ addition ----> multiplication power rule $\large\rm log_b x^y = y \log_b x$

17. Nnesha

heheh change of base formula $\huge\rm \log_b a = \frac{ \log a }{ \log b }$

18. Nnesha
19. anonymous

\begin{align} 2\log(100^{-3}) &=2\log\left(\frac{1}{100^3}\right) \\& = 2[\log(1) -\log(100^3)] \\ &=2[0-\log(10^6)] \\&= 2[-6\log(10)] \\&= 2(-6(1))\\&= -12 \end{align}

20. anonymous

That's another approach you could have used as well :P

21. marcelie

ah i see it :) Thank you ladies :)

22. anonymous

@Nnesha what do you think?

23. Nnesha

nice!

24. Nnesha

or after this if you don't want to use change of base formula set it equal to x $\huge\rm log (100^{-6}) =x$ now change log to an exponential form log= log base 10 $10^x =100^{-6}$ rewrite 100^{-16} in terms of base 10 $10^x =10^{2 \times -6}$ bases are same you can cancel $\cancel{10}^x =\cancel{10}^{2 \times -6}$ left with x= 2 times -6 = -12 :=)

25. Nnesha

|dw:1444012127376:dw| that's all you need to know for logs or u can visit that website :=)

26. anonymous

$\large \log_{10}(100^{-6}) = x$$10^{\large \log_{10}(100^{-6})} = 10^x$$\large 100^{-6} = 10^{x} \qquad \qquad \qquad 10^{-12} \equiv 100^{-6}$$\large 10^{-12} = 10^{x}$$\therefore -12 = x$