We have to find the solution of 5^(11x-1)=7^x

5^(11x-1)=7^x

=> 5^11x / 5 = 7^x

take the logarithms of both the sides

log( 5^11x / 5) = log 7^x

=> log 5^11x - log 5 = x log 7

=> 11x log 5 - log 5 = x log 7

=> 11x log 5 - x log 7 = log 5

=> x( 11*log 5 - log 7) = log 5

=> x = log 5 / ( 11*log 5 - log 7)

=> x = 0.102 approximately.

**Therefore x = log 5 / ( 11*log 5 - log 7)**

To solve for x in 5^(11x-1)=7^x.

We take the logarithms with respect to base 10 of both sides of the equation 5^(11x-1)=7^x.

(11x-1) log5 = x log7.

(11x)log5- log5 - xlog7 = 0.

(11x)log5 - xlog7 = log5.

(11log5 - log7)x = log5.

x = log5/{11log5 - log 7) = 0.1021 nearly.

Since the bases are not matching, we can use logarithms to solve exponential equations.

We'll take logarthims both sides:

log5 [5^(11x-1)] = log5 (7^x)

We'll apply the power rule for logarithms:

(11x-1) log5 5 = x log5 7

We'll recall that log5 5 = 1

We'll re-write the equation:

11x-1 = x log5 7

We'll subtract x log5 7 both sides:

11x - x log5 7 = 1

We'll factorize by x:

x(11 - log5 7) = 1

We'll re-write log5 7 = lg7/lg5

x(11 - lg7/lg5) = 1

We'll divide by 11 - lg7/lg5 = 9.7909

x = 1/9.7909

Rounded to four decimal places:

**x = 0.1021**