Basic Logs, simplifying So basically the question goes: $\log_{14} 2 = a$, $\log_{14} 3 = b$, solve for $\log_7 24$. I have attached my work so far and the answer is ${3a+b}\over{1-a}$... I just got stuck at the end. Thanks
 A: Using: $log_x y={\log_m y\over{\log_m x}}$, $\log_x y^c=c\log_x y$ and $\log_x a+\log_x b=\log_x ab$ we get 
$\log_{7} 24 = 3\log_7{2}+\log_7{3}={3\log_{14}2\over{\log_{14}7}}+{\log_{14}3\over{\log_{14}7}}={{3\log_{14}2+\log_{14}3}\over{\log_{14}7 }}$
$$={{3\log_{14}2+\log_{14}3}\over{\log_{14}7 +\log_{14}2-\log_{14}2}}$$
$$={{3\log_{14}2+\log_{14}3}\over{\log_{14}14-\log_{14}2}}={{3a+b}\over{1-a}}$$
A: In your approach you tackle the problem using the definition: $\log_x y = z \iff y=x^z.$ You just need to solve $\log_7 2$, and you'd be done. Jump to the second-to-last line for that.

We know that $\log_{14} 2 = a$ means $2=14^a$, and $\log_{14} 3 = b$ means $3=14^b$. 
We'd like to solve for $\log_{7} 24 = x.$ In other words, $24=7^x$.
Now notice that $2\cdot 2\cdot 2 \cdot 3= 24$.
Substituting what we know from our first line above, we have:
$$14^a 14^a 14^a 14^b = 14^{3a+b} = (2\cdot 7)^{3a+b} = \color{green}{2^{3a+b} 7^{3a+b} = 7^x}.$$
Taking $\log_7$ of both sides of the equation in green, we have 
$$(3a+b)\log_7 2 +(3a+b) = x.$$
So, $$(3a+b)(\log_7 2 + 1) = x.$$
But what is $\log_7 2$? Well, notice that $2=14^a=2^a7^a$, which is equivalent to $2^{1-a}=7^a$. So $(1-a)\log_7 2 = a$. Therefore $\log_7 2 = \dfrac{a}{1-a}$.
So, $$(3a+b)\left(\frac{a}{1-a} + 1\right) = x.$$
