Respuesta :
Answer:
The value of log 43is 1.633
Step-by-step explanation:
Explanation:
Suppose you know that:
[tex]\begin{array}{l}{\log 2 \approx 0.30103} \\{\log 3 \approx 0.47712}\end{array}[/tex]
Then note that:
[tex]43=\frac{129}{3} \approx \frac{128}{3}=\frac{2^{7}}{3}[/tex]
So
[tex]\log 43 \approx \log \left(\frac{2^{7}}{3}\right)=7 \log 2-\log 3 \approx 7 \cdot 0.30103-0.47712=1.63009[/tex]
We know that the error is approximately:
[tex]\log \left(\frac{129}{128}\right)=\log 1.0078125=\frac{\ln 1.0078125}{\ln 10} \approx 0.00782 .3=0.0034[/tex]
So we can confidently give the approximation:
[tex]\log 43 \approx 1.633[/tex]
Answer:
1.6
Step-by-step explanation:
It's 1.633, when you round to the nearest tenth, it's B) 1.6