Logarithm properties doubt

Question

The problem is $\log (5.64)^4$. According to the properties and laws of exponents, $\log (m^r) = r \log (m)$. But since the exponent is outside of the parenthesis in this problem, does it solves by like $4\log (5.64)$ or $(\log (5.64))^4$? TYIA.$ \ \ \ \ $

Answer 1

You're right that $\log(a^b)=b\log(a)$. Here, we have $(\log(a))^b$ indeed. This cannot be simplified easily. You could do the following, but I don't think that will help you: $$ (\log(a))^2=\log(a)\log(a)=\log(a^{\log a}) $$ To calculate the answer, just calculate $\log 5.64$ and take the fourth power of that.

Answer 2

@Ragnar has changed the statement of the problem. In the original problem, it is ambiguous. In the modified problem, it is clearly the case that the exponent is to be applied after the logarithm. $(\log 5.63)^4$ is not ambiguous, nor is $\log 5.63^4$ (in the second, the exponent is applied before the log). For that matter, $(\log(5.63))^4$ is not ambiguous. But $\log(5.63)^4$, which is how the problem was originally stated, is ambiguous, since it isn't clear whether you intend to treat the function as $\log$, with no parentheses (like $\sin$), or as $f(\cdot)$ like generic function names.

Answer 3

Your answer will not follow the law you have given. To see the difference, notice that $\log(5.63)$ 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 $$(10^a)^4=5.63^4$$

Applying exponential rules gets $$10^{4a}=5.63^4$$

But remember that we let $a=\log(5.63)$. 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 $4*a$ gets $5.63^4.$ That is, $$\log(5.63^4)=4a=4*\log(5.63)$$

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 $(\log(5.63))^4$. Here, following order of operations, you must first calculate $(\log(5.63)$ 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 $a=\log(5.63)$, your solution is $a^4$ instead of $4a$.