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I have two practice problems I am working through and I am stuck as to the logic involved in solving them.

  1. $100^{\log(5/2)}$
  2. $100^{\log(5)/2}$

What properties should I be looking at here and if you could detail the steps involved it would be much appreciated. Thanks.

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closed as off-topic by TheGeekGreek, user91500, user21820, Namaste, Siong Thye Goh Aug 27 '17 at 14:28

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    $\begingroup$ What are we supposed to do with them? $\endgroup$ – Parcly Taxel Aug 27 '17 at 3:03
  • $\begingroup$ We are supposed to get a numerical value. I already know that the answers are 6.25 and 5, I am just unsure as to the steps involved to get there. $\endgroup$ – xCargot Aug 27 '17 at 3:26
  • $\begingroup$ No calculators? Are you given $\log5$ and $\log2$? $\endgroup$ – Parcly Taxel Aug 27 '17 at 3:26
  • $\begingroup$ @RyePie The first question can be written as $100 \log 5 - 100 \log 2$, which surely is much more than $6.25$. How did you calculate your answer? $\endgroup$ – Toby Mak Aug 27 '17 at 3:31
  • $\begingroup$ Sorry, I entered the problem incorrectly. I didn't have the logs set to an exponent in either equation. It is correct now. $\endgroup$ – xCargot Aug 27 '17 at 3:49
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Assuming that $\log$ is the base-$10$ logarithm and using $10^{\log(x)} = x$ for all $x \in \mathbb{R}_{>0}$, we obtain $$ \begin{array}{lrcccl} & 100^{\log(5/2)} &=& \big( 10 \cdot 10 \big)^{\log(5/2)} = 10^{\log(5/2)} \cdot 10^{\log(5/2)} = \dfrac{5}{2} \cdot \dfrac{5}{2} = \dfrac{25}{4} &=& 6.25 \\ \text{and} \\ & 100^{\log(5)/2} &=& \big( 10 \cdot 10 \big)^{\log(5)/2} = \big( 10^{\log(5)} \cdot 10^{\log(5)} \big)^{1/2} = \big( 5 \cdot 5 \big)^{1/2} &=& 5 \end{array} $$

The last one can also be computed as follows: $$ \begin{array}{rcccl} 100^{\log(5)/2} &=& \big( 10^2 \big)^{\log(5)/2} = \big( 10^{2/2} \big)^{\log(5)} = 10^{\log(5)} &=& 5 \end{array} $$

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  • $\begingroup$ Did you mean $10^{\log x} = x$ for all $x \in \mathbb{R}^+$? $\endgroup$ – N. F. Taussig Aug 27 '17 at 7:25
  • $\begingroup$ @N.F.Taussig: Yes, thank you. I edi it. $\endgroup$ – Björn Friedrich Aug 27 '17 at 7:26
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$1.$ Let $a = 100^{\log(5/2)}$

Taking $\log$ on both sides (base $10$), we finally have $$2(\log 5 - \log 2) = \log a$$ $$\implies 0.7958 = \log a$$ $$\implies a = 6.5$$(approx.)

Similarly you can solve the second one by taking $\log$ on both sides, and you finally get the answer i.e. $5$

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Remember, $10^{log(x)}$=x. (the base of log(x) is 10). This is simply because $log(x)$ is saying "What power do we need to raise 10 to in order to get x?" Raising 10 to that very power will obviously get us x.

Right here, you have $100^{log(x)}$. Remember that $100=10^{2}$.

$(10^{2})^{log(x)}$ is simply $10^{2log(x)}$. You should know this from exponent rules and just by conceptualizing what happens.

(To give an example, $(3^{4})^{2}=3^{4} \times 3^{4}$ which, as you know by exponent rules, you add the powers of 4, which just means multiplying by 2.)

Use the fact that $100^{log(x)}$=$(10^{2})^{log(x)}$=$10^{2log(x)}$ to solve the problems easily. You may also take advantage of the fact that $2log(x)=log(x^{2})$ to really see how the complicated-looking exponents reduce down to something simple.

Bottom line: review your log rules, and get some practice using them. Then you will automatically see how to make a problem easier by using log rules.

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