How do you evaluate ${log}_{\sqrt{7}} 49$ $log_sqrt7 49$ ?

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Answer 1 · 2016-04-04T11:33:37

$4$ $4$

We should attempt to write $49$ $49$ as a power of $\sqrt{7}$ $sqrt7$ .

First, simply rewrite $49$ $49$ as $7^{2}$ $7^2$ .

${log}_{\sqrt{7}} 49 = {log}_{\sqrt{7}} (7^{2})$ $log_sqrt7 49=log_sqrt7(7^2)$

This can be rewritten using the rule:

${log}_{a} (b^{c}) = c \cdot {log}_{a} b$ $log_a(b^c)=c*log_ab$

Thus, we have

${log}_{\sqrt{7}} (7^{2}) = 2 {log}_{\sqrt{7}} 7$ $log_sqrt7(7^2)=2log_sqrt7 7$

Now, we can write that $7 = {(\sqrt{7})}^{2}$ $7=(sqrt7)^2$ .

$2 {log}_{\sqrt{7}} {(\sqrt{7})}^{2}$ $2log_sqrt7(sqrt7)^2$

Rewrite using the rule we found earlier.

$2 \cdot 2 {log}_{\sqrt{7}} \sqrt{7}$ $2*2log_sqrt7sqrt7$

Note that

${log}_{a} a = 1$ $log_aa=1$

So we are just left with

$2 \cdot 2 {log}_{\sqrt{7}} \sqrt{7} = 4$ $2*2log_sqrt7sqrt7=4$

We can test this by using our rules of logarithms. We said that:

${log}_{\sqrt{7}} 49 = 4$ $log_sqrt7 49=4$

This means that ${(\sqrt{7})}^{4} = 49$ $(sqrt7)^4=49$

$\sqrt{7} \times \sqrt{7} \times \sqrt{7} \times \sqrt{7} = 7 \times 7 = 49$ $sqrt7 xx sqrt7xxsqrt7xxsqrt7=7xx7=49$

How do you evaluate log√749log_sqrt7 49?