# Solving three exponential equations simultaneously

I am doing a math project and I was wondering how I should approach finding the coefficients of these three exponential functions. I thought of solving them simultaneously, but I didn't think it would work.

$$ab^{2c} = 3674140$$

$$ab^{3c} = 4.325200327*10^{19}$$

$$ab^{4c} = 7.40119684×10^{45}$$

I am having trouble because I know there isn't a formula that goes through all the points, but relatively follows the path of them. Please let me know what method I should use to find the coefficients a, b, and c.

• By the way, the numbers 2, 3, and 4 mean when x = 2, 3 and 4. – Aryaman D. Sep 13 at 14:27
• If you apply $\log$ to both sides of each equation, you get a system of three linear equations in $\log a$, $\log b$, and $c$. (It may be helpful to write $A$ for $\log a$ and $B$ for $\log b$ when solving, to reduce clutter.) Is that helpful? – MPW Sep 13 at 14:29
• the system is not consistent. – Will Jagy Sep 13 at 17:42
• As @Will Jagy remarked, the system is contradictory : if we denote by (1), (2), (3) the 3 equations, the quotient (2)/(1) gives $b^c=1.77*10^{13}$ whereas the quotient (3)/(2) gives $b^c=1.71 * 10^{26}$... for the same $b^c$... But maybe you wanted to place parentheses around $ab$ : $(ab)^{2c}=...$ etc. ? – Jean Marie Sep 13 at 21:12
• @JeanMarie is there any way to find a function that goes near the points of defines the relationship? I know these points aren't consistent, but I just need to find an equation that goes near. – Aryaman D. Sep 14 at 5:58

Denoting the right-hand sides by $$A,B,C$$ then we get with $$A=\frac{A}{b^{2c}}$$ $$b^c=\frac{B}{A}$$ and $$b^{2c}=\frac{B}{A}$$ Can you finish? So $$b^c=b^{2c}$$