Question

$5^{x-5} + 5^{4-x}=\frac{6}{5}$

Answer 1

To Solve:

           $5^{x-5}+5^{4-x}=\frac{6}{5}$

Solution:

we know that ,

             $a^m+a^n=a^{m+n}$

so,

  $5^{(x-5)(4-x)}=\frac{6}{5}\\\\5^{4x-x^2+5x-20}=\frac{6}{5}\\\\5^{x^2+9x-20}=\frac{6}{5}$

             

Answer 2

$\large\text{\underline{Question}}$

What value of $x$ satisfies the exponential equation?

Given exponential equation.

$\red{\bullet}\ 5^{x-5}+5^{4-x}=\dfrac{6}{5}$

$\large\text{\underline{Step 1. Exponent law}}$

We know that $5^{x-5}=\dfrac{5^{x}}{5^{5}}$ and $5^{4-x}=\dfrac{5^{4}}{5^{x}}$.

$\large\text{\underline{Step 2. Rational equation}}$

For the sake of clarity, we will show $5^{x}$ as $t$.

Given rational equation.

$\red{\bullet}\ \dfrac{t}{5^{5}} +\dfrac{5^{4}}{t}=\dfrac{6}{5}$

To resolve this problem we will multiply the same numbers. In this case, multiplying by $5^{5}\times t$ will resolve the problem.

$\implies t^{2}+5^{9}=6\cdot5^{4}\cdot t$

$\implies t^{2}-6\cdot5^{4}\cdot t+5^{9}=0$

So, now we can solve this by factorization.

$\implies (t-5^{4})(t-5^{5})=0$

$\implies t=5^{4}\text{ or }t=5^{5}$

$\large\underline{\text{Step 3. With respect to }x}$

So, now we have to solve for $x$. We know that $t=5^{x}$.

$\implies5^{x}=5^{4}\text{ or }5^{x}=5^{5}$

The equation has $x=4,5$ as solutions.

$\large\underline{\text{Conclusion}}$

The value of $x$ are $x=4,5$.