How to inverse log of an expression

Question

I have a formula $$ \log_{10}C= (\log_{10}A - 0.80 \cdot \log_{10}B - 8.40)/0.50 $$ I want to write this down as non-logaritmic. I cannot be sure about the $-0.80\cdot \log_{10}B$ part It might be one of these:

$$ C= (A - 10^{0.80} \cdot B - 10^{8.40})/10^{0.50} $$ or $$ C= (A - B^{0.80} - 10^{8.40})/10^{0.50} $$

Am I correct or totally wrong is there any other correct one?

Answer 1

To note, it seems like your knowledge of logarithms and exponents is a bit chaotic.

Let's recall the following properties of exponents and logarithms, $\forall a,b,c \in \mathbb{R}_+$:

1. $b \neq 1:~ a = b^{\log_b a}$.
2. $a^b \cdot a^c = a^{b+c}$ and $a^b \cdot a^{-c} = a^{b-c}$.
3. $(a^b)^c=(a^c)^b = a^{bc}$.
4. $a\neq 1:~\log_a(bc) = \log_a b+\log_a c$, and $\log_a (b/c) = \log_a b - \log_a c$.
5. $a\neq 1:~\log_a b^c = c\log_a b$.
6. If $a=b$ then $c^a = c^b$.

Knowing the above, we can easily show that: $$ \log_{10}C = 2\log_{10} A-1.6\log_{10} B -16.8 \\ 10^{\log_{10}C} = 10^{2\log_{10} A-1.6\log_{10} B -16.8} \\ C = \dfrac{10^{\log_{10}A^2}}{10^{\log_{10}B^{1.6}}\cdot 10^{16.8}} \\ C = \dfrac{A^2}{10^{16.8}\cdot B^{1.6}} $$ Hope my explanation was clear enough, if you still have any questions, please don't hesitate to ask them in the comment section :)