Shouldn't the divisor be a prime number?

Since 15 and 26 are coprimes, we can apply Euler's theorem , which means $15^{\phi (26)} \equiv 1 \text{ (mod 26)}$ , where $\phi(n)$ is the Euler's totient function , and $\phi(26) = 26\left(1-\dfrac{1}{2}\right) \left(1-\dfrac{1}{13}\right) = 12$ . Therefore, we have:

$\begin{aligned} 15^{145} & \equiv 15^{145 \text{ mod } \phi (26)}\text{ (mod 26)} \\ & \equiv 15^{145 \text{ mod } 24}\text{ (mod 26)} \\ & \equiv 15^1 \text{ (mod 26)} \\ & = \boxed{15} \text{ (mod 26)} \end{aligned}$

Observe that $15^{145}\equiv1\bmod2$ . Also observe that $15^{145}\equiv2^{145}\bmod13\equiv(2^6)^{24}*2\bmod13\equiv2\bmod13$ . Then, using Chinese Remainder Theorem, we have 15 mod 26 as the answer.

log with base 15 ( x )= 26 then x =1.205 ## ### 15^145 /15^143.785= 15^1,..... thats bad soultion but true