# Logarithm properties doubt The problem is log &#x2061;<!-- ⁡ --> ( 5.64 )

Logarithm properties doubt
The problem is $\mathrm{log}\left(5.64{\right)}^{4}$. According to the properties and laws of exponents,
$\mathrm{log}\left({m}^{r}\right)=r\mathrm{log}\left(m\right)$. But since the exponent is outside of the parenthesis in this problem, does it solves by like $4\mathrm{log}\left(5.64\right)$ or $\left(\mathrm{log}\left(5.64\right){\right)}^{4}$? TYIA.
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Sariah Glover
Your answer will not follow the law you have given. To see the difference, notice that $\mathrm{log}\left(5.63\right)$ represents the power of 10 needed to attain a result of $5.63$, i.e. it is the number (call it $a$ for short) that fits the equation
${10}^{a}=5.63$
Raising both sides to the fourth power gets
$\left({10}^{a}{\right)}^{4}={5.63}^{4}$
Applying exponential rules gets
${10}^{4a}={5.63}^{4}$
But remember that we let $a=\mathrm{log}\left(5.63\right)$. It follows from the above equation that, if raising $10$ to the power of $a$ gets $5.63$, then raising $10$ to the power of $4\ast a$ gets ${5.63}^{4}.$ That is,
$\mathrm{log}\left({5.63}^{4}\right)=4a=4\ast \mathrm{log}\left(5.63\right)$
Think about how this is different from your question, and you should clearly see why this rule won't apply for you. Your question is to solve $\left(\mathrm{log}\left(5.63\right){\right)}^{4}$. Here, following order of operations, you must first calculate $\left(\mathrm{log}\left(5.63\right)$ and then raise it to the fourth power, rather than first considering $5.63$ to the fourth power and then calculating logarithm as we did above. That is to say, using our earlier substitution for $a=\mathrm{log}\left(5.63\right)$, your solution is ${a}^{4}$ instead of $4a$
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prirodnogbk
You're right that $\mathrm{log}\left({a}^{b}\right)=b\mathrm{log}\left(a\right)$. Here, we have $\left(\mathrm{log}\left(a\right){\right)}^{b}$ indeed. This cannot be simplified easily. You could do the following, but I don't think that will help you:
$\left(\mathrm{log}\left(a\right){\right)}^{2}=\mathrm{log}\left(a\right)\mathrm{log}\left(a\right)=\mathrm{log}\left({a}^{\mathrm{log}a}\right)$
To calculate the answer, just calculate $\mathrm{log}5.64$ and take the fourth power of that.