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