2loga x=1+loga (7x-10a)?

Question

find x in terms of a.
answer 2a,5a

Answer 1

loga (x²) = 1 + loga (7x - 10a)
loga (x²) - loga (7x - 10a) = 1
loga (x² / (7x - 10a) ) = loga (a)
x² / (7x - 10a) = a
x² = 7a.x - 10a²
x² - 7.ax + 10a² = 0
(x - 5a).(x - 2a) = 0
x = 5a , x = 2a

Answer 2

2loga x = 1 + loga (7x - 10a)

loga (x^2) = loga a + loga (7x - 10a)

loga ((x^2)/(a*(7x-10a)) = 0

(x^2)/(a*(7x-10a)) = 1

x^2 = 7ax - 10a^2

x^2 - 7ax + 10a^2 = 0

Quadratic A=1, B= -7a, C= 10a^2 and solve

Answer 3

from that equatation you can change to
loga x^2=loga a+loga(7x-10a)
cuz loga a=1,
use product rule for log
loga x^2=loga a(7x-10a)
both sides they have the same log withg base a
therefore
x^2=a(7x-10a)
x^2=7ax-10a^2
x^2-7ax+10a^2=0
this is quaratic formular
a=1, b=-7a, c=10a^2
x= (7a+/- sqrt(49a^2-40a^2))/2
you just replace the a,b,c number
x=(7a+/-sqrt(9a^2))/2
x=(7a+/- 3a)/2
therefore
x=(7a+3a)/2= 5a
x=(7a-3a)/2=2a
that is what they got. good luck