$a, b, c, d \in \mathbb{R}^+$, $\log_a(b) = 8/9, \log_b(c) = -3/4$ and $\log_c(d) = 2$, find the value of $\log_d(abc)$

Question

I am doing my homework and I came across this question that is pretty difficult for me. I do not know how to advance after this certain step. The question reads, "If $a, b, c$ and $d$ are positive real numbers such that $\log_a(b) = 8/9, \log_b(c) = -3/4$ and $\log_c(d) = 2$, find the value of $\log_d(abc)$."

I have converted all of the following into their exponential forms, but I do not know how to find the values from there. I am sure if one value is found, the rest can be found easily from there.

If anyone can offer their guidance, that would be greatly appreciated.

Answer 1

Hint: $\log_d(abc)=\log_da+\log_db+\log_dc$. If we can get those three terms in the RHS, our question is solved. Now, we are also given three equations:

$$\log_ab=8/9...(1)$$ $$\log_bc=-3/4...(2)$$ $$\log_cd=2...(3)$$

Multiply $(2)$ and $(3)$, the $c$ will cancel out. What are you left with?
Multiply $(1)$, $(2)$ and $(3)$, the $b$ and $c$ will cancel out. What are you left with?

Do you now have all the three terms you originally wished to compute?

Answer 2

From the definition of logarithms, you have $$\log_{a}(b) = 8/9 \implies b = a^{8/9} \tag{1}$$ $$\log_{b}(c) = -3/4 \implies c = b^{-3/4} \tag{2}$$ $$\log_{c}(d) = 2 \implies d = c^{2} \tag{3}$$

Now, \begin{align} \log_{d}(abc) = \log_{d}(a) + \log_{d}(b) + \log_{d}(c) \tag{4} \end{align}

From (3), $$d = c^{2} \implies d^{1/2} = c \implies \log_{d}(c) = 1/2.$$ Similarly, from (2) and (3) we have $$d = c^{2} = (b^{-3/4})^{2} = b^{-3/2} \implies d^{-2/3} = b \implies \log_{d}(b) = -2/3.$$

Following the same analogy, from (1) and (2), $$d = b^{-3/2} = (a^{8/9})^{-3/2} = a^{-4/3} \implies d^{-3/4} = a \implies \log_{d}(a) = -3/4.$$

I hope you can solve the rest.

Answer 3

Change of base gives the following ratios \begin{eqnarray*} \log_a b =\frac{8}{9} = \frac{\ln(b)}{\ln(a)} \\ \log_b c =-\frac{3}{4} = \frac{\ln(c)}{\ln(b)} \\ \log_c d =2 = \frac{\ln(d)}{\ln(c)}. \\ \end{eqnarray*} From these ratios we have the following ratios \begin{eqnarray*} \frac{\ln(c)}{\ln(d)} =\frac{1}{2}\\ \frac{\ln(b)}{\ln(d)} =-\frac{2}{3}\\ \frac{\ln(a)}{\ln(d)} =-\frac{3}{4}\\ \end{eqnarray*} So \begin{eqnarray*} \log_d (abc) =\frac{\ln(abc)}{\ln(d)} =\frac{\ln(c)}{\ln(d)} +\frac{\ln(b)}{\ln(d)} +\frac{\ln(a)}{\ln(d)} =\frac{1}{2}+-\frac{2}{3}+-\frac{3}{4}=? \end{eqnarray*}