√x^(log(x−1)÷log(5))=5

x=25

Step-by-step explanation:

The actual question is shown in the image attached.

Solve for x.

sqrt(x)^(log_5(x) -1) = 5  ......................(1)

Solution:

from (1)

sqrt(x)^(log_5(x) -1) = 5  .......separate exponents, law of log a^(b-c)=a^b/a^c

sqrt(x)^log_5(x) / sqrt(x) = 5  .......... cross multiply

sqrt(x)^log_5(x) = 5sqrt(x)   .............. square both sides

(sqrt(x)^log_5(x))^2 = 25x     .............. modify base a^(2b) = (a^2)^b

(sqrt(x)^2)^log_5(x) = 25x

x^log_5(x) = 25x   .................... take log_5 on both sides

log_5(x) * log_5(x) = log_5(5^2*x)  ............... simplify RHS

log_5(x) * log_5(x) = log_5(25)+log_5(x)

log_5(x) * log_5(x) = 2+log_5(x) ........ simplify

log_5(x) ^2 -log_5(x) -2 = 0   ........... substitute y = log_5(x)

y^2 - y -2 = 0

(y-2)(y+1) = 0

y=2 or

y = -1   ................... y = log_5(x) >= 0 , y=-1 rejected

y = 2

log_5(x) = 2

raise to base of 5

5^log_5(x) = 5^2

x = 25

Check by substituting x = 25 in (1)

sqrt(x)^(log_5(x) -1)

= sqrt(25)^(log_5(25) -1)

= 5^(2-1)

= 5    equal RHS,  therefore solution is correct.

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1