I've just been accepted on to a PHD program at Melbourne, studying chemical engineering.

I'm working my way through some standard pure and further mathematics books just to get the concepts into my head before I start. Some of this I won't need, but maths is invaluable in the sciences anyway.

I've been going through some logs and exponentials questions. I've found that I'm doing really well, i've just come across one set of questions and between flicking back and forth from the laws of logs and exponentials and the questions themselves, I'm getting a bit frustrated.

The textbook is asking me to find $a$ if:

1) $\log_a 8 = \frac 3 2$

2) $\log_a 25 - \log_a 0.25 = 2$

3) $\log_2 a - \log_8 a = 4$

In terms of doing the same to both sides, where is the most logical place to start? Clearly for the first it is to multiply both sides by $2$ so I have $2 \log_a 8 = 3$.

  • $\begingroup$ The inverse operation to taking the $\log_a(n)$ is exponentiating with the base $a$. So we would have $\log_a(8) = 3/2 \Longleftrightarrow 8 = a^{3/2}$. Also $a^{\log_a(x)} = x$, just like $e^{\ln(x)} = x$ $\endgroup$ – Maximilian Gerhardt May 13 '16 at 9:52

Hint $1$: $\log_{a}{b}=x \Rightarrow a^x=b$

Hint $2$: $\log_{a}{b} - \log_{a}{c} = \log_{a}(\frac{b}{c})$

Hint $3$: $\log_{a^k}{b} =(\frac{1}{k}) \log_{a}{b}$

Can you take it from here?

Since you didn't reply, looking at the history of your questions, I decided to give you the solution in as much detail as possible. Hope it helps.

$1)$ $\log_a 8 = \frac{3}{2}$

Using Hint $1$, we get $a^{\frac{3}{2}}=8 \Rightarrow a=8^{\frac{2}{3}}=4$

$2)$ $\log_a 25 - \log_a 0.25 = 2$

Using Hint $2$, we have $\log_{a}{(\frac{25}{0.25})}=\log_{a}{100}=2$

Now, using Hint $1$, we get $a^2=100 \Rightarrow a=(100)^{\frac{1}{2}}=10$

$3)$ $\log_2 a - \log_8 a = 4$

Using Hint $3$, we have $\log_8 a =(\frac{1}{3}) \log_2 a$

So, the given expression is

$(\frac{2}{3}) \log_2 a=4 \Rightarrow \log_2 a=4 (\frac{3}{2})=6$

Now, using Hint $1$, we get $a=2^6=64$


To solve $\log_xa=y$ for $a$, take the base $x$ antilogarithm


To solve $\log_ax=y$, use

$$\log_xa=\frac1{\log_ax}=\frac1y$$ and take the base $x$ antilogarithm


For more "complex" equations, try to transform in one of the above cases.


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