Problem: If log2=.3010\log 2=.3010log2=.3010 and log3=.4771\log 3=.4771log3=.4771, the value of xxx when 3x+3=1353^{x+3}=1353x+3=135 is approximately:
Answer Choices:
A. 555
B. 1.471.471.47
C. 1.671.671.67
D. 1.781.781.78
E. 1.631.631.63 Solution:
3x⋅33=135;∴3x=53^{x} \cdot 3^{3}=135 ; \quad \therefore 3^{x}=53x⋅33=135;∴3x=5;
∴xlog3=log5=log(10/2)=log10−log2\therefore x \log 3=\log 5=\log (10 / 2)=\log 10-\log 2∴xlog3=log5=log(10/2)=log10−log2.
∴x=(1−0.3010)/0.4771∼1.47\therefore x=(1-0.3010) / 0.4771 \sim 1.47∴x=(1−0.3010)/0.4771∼1.47.