Problem: If (0.2)x=2(0.2)^{x}=2(0.2)x=2 and logβ‘2=0.3010\log 2=0.3010log2=0.3010, then the value of xxx to the nearest tenth is:
Answer Choices:
A. β10-10β10
B. β0.5-0.5β0.5
C. β0.4-0.4β0.4
D. β0.2-0.2β0.2
E. 101010 Solution:
(210)x=2;β΄logβ‘(210)x=xlogβ‘210=x(logβ‘2βlogβ‘10)=logβ‘2\left(\dfrac{2}{10}\right)^{x}=2 ; \quad \therefore \log \left(\dfrac{2}{10}\right)^{x}=x \log \dfrac{2}{10}=x(\log 2-\log 10)=\log 2(102β)x=2;β΄log(102β)x=xlog102β=x(log2βlog10)=log2.
β΄x=0.30100.3010β1βΌβ0.4\therefore x=\dfrac{0.3010}{0.3010-1} \sim-0.4β΄x=0.3010β10.3010ββΌβ0.4.