Logarithms with an answer that is a fraction

Question

How does log base $16$ of $32$ equal $1.25$? If we divide $32/16=2$ but then if we divide $2/16$ it doesn't come out to a whole number unlike with log base $2$ of $4$ where $4/2=2$ and $2/2=1$

I am confused why its $1.25$ and not simply $1$? If we further try to divide with $2/16$ we get $.125$ giving us a total of $1.125$ not $1.25$. However since $2$ can't be divided by $16$ it doesn't make sense to keep going.

Obviously the answer is $1.25$ since $16^{1.25}=32$, I'm just not sure how we can get that number.

Answer 1

We have that:

$$\log_{16}{32} = 1.25 = \frac{5}{4}$$

Notice that this can also be represented as: $16^{5/4} = 32$

Why? Well, first represent $16$ as $2^4$

It then follows that: $16^{5/4} = (2^4)^{5/4} = 2^5 = 32$

Answer 2

$\log_{16} 32=1.25$ because $16^{1.25}=32$

Recall that in general,

$\log_{a} b=c$ means that $a^{c}=b$

Addendum:

If you are asking how to determine what $\log_{16}32$ is, we first change it into exponent form as follows:

$\log_{16}32=x$ means $16^x=32$. We then take the log of both sides.

$\log 16^x=\log 32$

Then using the property of logs that lets us move the exponent down (i.e, $\log x^n=n\log x)$, we can simplify this to

$x\log 16=\log32\implies x=\frac{\log32}{\log 16}=1.25$

Answer 3

In general, $\log_{a^n}(a^m) = \frac{m}{n}$.

One way to see this result is the general rule:

$$\log_x y = \frac{\log_a y}{\log_a x}$$

I your case, $a=2$, $m=5$ and $n=4$.

This is the only way for $\log_{x}y $ to be rational if $x,y$ are integers greater than $1$. This is because if $x=y^{p/q}$, with $p,q$ relatively prime integers, then let $M=x^{q}=y^p$, and you can use unique factorization to show that the powers of the primes in the prime factorization of $M$ must be divisible by both $p$ and $q$, and hence by $pq$. That means there is an integer $a$ so that $M=a^{pq}$, and $a^{q}=x$ and $a^p=y$.

Answer 4

In elementary school we were initially taught that multiplication is just repeated addition $$3\times5=3+3+3+3+3$$ Then as we were introduced to fractions, we learned that a fraction is just repeated addition of a sector fraction $$\frac{3}{5}=3\times\frac{1}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$$ Then we were asked to consider $$\frac{1}{3}\times\frac{1}{5}$$ So we reasoned "I need to add $1/3$, $1/5$ times.", "How do we add something $1/5$ times?", "What does that mean?", "Teacher ...... HELP!". We eventually dropped the "repeated addition" interpretation and began it interpret such expressions as proportions $$\frac{1}{3}\times\frac{1}{5}\iff \frac{1}{3}\text{ of }\frac{1}{5}$$ We did this by appealing to the other properties of multiplication and addition $$3\times5\times\frac{1}{3}\times\frac{1}{5}=3\times\frac{1}{3}\times\frac{1}{5}\times 5=(3\times\frac{1}{3})\times(\frac{1}{5}\times 5)=1\times1=1$$ In this case, we used the commutative and associative properties of multiplication to show that $1$ is $15$ times larger than $\frac{1}{3}\times\frac{1}{5}$, implying that $$\frac{1}{3}\times\frac{1}{5}=\frac{1}{3\times 5}=\frac{1}{15}$$ Later we learned about exponentiation, and were told that exponentiation is just repeated multiplication $$2^4=2\times2\times2\times2$$ and we interpreted the above expression as $2$ multiplied $4$ times. Then came another "Teacher.....HELP!" moment when we were asked to consider $$16^\frac{1}{4}$$ and we asked ourselves "How can I multiply $16$ one quarter times? How can I multiply anything a fractional number of times? I don't even know what that means."

Again, we deferred to the other properties of exponentiation. $$16^\frac{1}{4}\times16^\frac{1}{4}\times16^\frac{1}{4}\times16^\frac{1}{4}=16^{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}}=16^1=16$$ but $$16^\frac{1}{4}\times16^\frac{1}{4}\times16^\frac{1}{4}\times16^\frac{1}{4}=(16^\frac{1}{4})^4$$ so it stands to reason that $$(16^\frac{1}{4})^4=16\iff 16^\frac{1}{4}=\sqrt[4]{16}=2$$

Applying all of this to your problem, $$32 = 16\cdot 2=16^1\cdot \sqrt[4]{16}=16^1\cdot 16^\frac{1}{4}=16^{1+\frac{1}{4}}=16^{1.25}$$ which, of course, is equivalent to $$\log_{16}32=1.25$$

Answer 5

So, let's say you want to take $x = \log_{16}32$, just like in your problem statement. However, you can't nicely write $32$ as an integer power of $16$. So, let's find the next power of $16$ that's also a power of $32$.

$$16 = 2^4$$

We need a number that's a power of $2^4$, more generally a number $2^{4n}$

$ 32 = 2^5 \\ 32^2 = 2^{10} \\ 32^3 = 2^{15} \\ 32^4 = 2^{20} = (2^4)^5 = 16^5 \leftarrow \text{This power of two is divisible by 4.}\\ $

So, instead of working with $\log_{16} 32$ we're going to work with $\log_{16} 32^4$.

Using log rules, we have:

$$\log_{16} 32^4 = 4 \log_{16} 32 = 4x$$

So we can rewrite the whole problem as:

$$\log_{16} 32^4 = \log_{16} 16^5 = 5 \log_{16} 16 = 5 = 4x$$

Solve the simple equation $5 = 4x$ and you have your answer of $x = 1.25$.

Answer 6

here is an answer to your question why $\log_{16} 32 = 1.25$ and not $1.0$

first we have $$\cdots, log_{16}( \dfrac{1}{16}) = -1, log_{16} (1) = 0, log_{16} (16) = 1, log_{16}( 256) = 2, \cdots$$ these can be written also in the exponential form as $$\cdots, 16^{-1} = \dfrac{1}{16}, 16^0 = 1,16^1 = 16, 16^2 = 256, \cdots$$ you can see the correspondence between the logarithm to the base $16$ of a number and the exponent to which the base $16$ must be raised so that it will the said number.

if we want to find the $log_{16}(32)$ you can guess that it must be between $1$ and $2$ because $16^1 = 1, 16^2 = 256$ and we are looking for the number $k$ so that $16^k = 32$ this is easier to handle because $16 = 2^4$ and $32 = 2^5.$ we can rewrite $16^k = 32$ as $(2^4)^k = 2^{4k} = 2^5 = 32$ so $4k = 5$ and $k = 1.25 $

Answer 7

The key idea here is to express both $16$ and $32$ as powers of the same base. For a base $b > 0$, with $b \neq 1$, the equation $b^u = b^v$ is only satisfied when the exponents are equal.
\begin{align*} \log_{16} 32 & = x\\ 16^x & = 32 && \text{by definition of the logarithm}\\ (2^4)^x & = 2^5 && \text{express each base as a power of $2$}\\ 2^{4x} & = 2^5 && \text{since $(b^m)^n = b^{mn}$}\\ 4x & = 5 && \text{since $b^u = b^v \Longleftrightarrow u = v$ if $b > 0$, $b \neq 1$}\\ x & = \frac{5}{4} \end{align*}