How do you evaluate log_49 7?

Jul 27, 2016

${\log}_{49} 7 = \frac{1}{2}$

Explanation:

Written in this form, the question being asked is.."How can I get 7 from 49?"

The answer is, by finding the square root. Another way of saying "square root" is the power of $\frac{1}{2}$.

In maths: ${\log}_{49} = x \text{ } \Rightarrow {49}^{x} = 7$

${49}^{\frac{1}{2}} = 7$

Therefore ${\log}_{49} 7 = \frac{1}{2}$

We can also ask "Which power of 49 is equal to 7?:

Jul 27, 2016

$\frac{1}{2}$

Explanation:

${\log}_{a} b = \frac{{\log}_{e} b}{\log} _ e a$

then

${\log}_{49} 7 = \frac{{\log}_{e} 7}{{\log}_{e} {7}^{2}} = \frac{{\log}_{e} 7}{2 {\log}_{e} 7} = \frac{1}{2}$

Jul 27, 2016

Have a look at https://socratic.org/s/awwES2YN

$x = 0.5$

Explanation:

$\textcolor{b l u e}{\text{Derived by calculation}}$

Set ${\log}_{49} \left(7\right) = x$

Write as ${49}^{x} = 7$.........................Equation(1)

Take logs to base 10 of each side of Equation(1)

${\log}_{10} \left({49}^{x}\right) = {\log}_{10} \left(7\right)$

This is the same as:

$x {\log}_{10} \left(49\right) = {\log}_{10} \left(7\right)$

$x = {\log}_{10} \frac{7}{\log} _ 10 \left(49\right)$

$x = 0.5$

Jul 27, 2016

$\frac{1}{2}$

Explanation:

Let:

$x = {\log}_{49} 7$

By the definition of logarithms, this can be rewritten as:

${49}^{x} = 7$

Write $49$ as ${7}^{2}$:

${\left({7}^{2}\right)}^{x} = 7$

Simplify ${\left({7}^{2}\right)}^{x}$ using the rule ${\left({a}^{b}\right)}^{b} = {a}^{b c}$:

${7}^{2 x} = 7$

Recall that $7 = {7}^{1}$:

${7}^{2 x} = {7}^{1}$

Since the powers are equal, we know their bases also must be equal:

$2 x = 1$

$x = \frac{1}{2}$