How do you use the Change of Base Formula and a calculator to evaluate the logarithm log_5 7?

Mar 4, 2018

${\log}_{5} \left(7\right) \approx 1.21$

Explanation:

The change of base formula says that:
${\log}_{\alpha} \left(x\right) = {\log}_{\beta} \frac{x}{\log} _ \beta \left(\alpha\right)$

In this case, I will switch the base from $5$ to $e$, since ${\log}_{e}$ (or more commonly $\ln$) is present on most calculators. Using the formula, we get:

${\log}_{5} \left(7\right) = \ln \frac{7}{\ln} \left(5\right)$

Plugging this into a calculator, we get:

${\log}_{5} \left(7\right) \approx 1.21$

Mar 4, 2018

$\text{Approx. } 1.209$.

Explanation:

The Change of Base Formula : ${\log}_{b} a = {\log}_{c} \frac{a}{\log} _ c b$.

$\therefore {\log}_{5} 7 = {\log}_{10} \frac{7}{\log} _ 10 5$,

$= \frac{0.8451}{0.6990} \approx 1.209$.

Mar 4, 2018

${\log}_{5} 7 \approx 1.21 \text{ to 2 dec. places}$

Explanation:

$\text{the "color(blue)"change of base formula}$ is.

â€¢color(white)(x)log_b x=(log_c x)/(log_c b)

$\text{log to base 10 just log and log to base e just ln}$
$\text{are both available on a calculator so either will}$
$\text{give the result}$

$\Rightarrow {\log}_{5} 7 = \frac{\log 7}{\log 5} \approx 1.21 \text{ to 2 dec. places}$

$\text{you should check using ln}$