how do you solve this logrithmic equation?

Question

log (base5)(x-1) +log(base5) x= log (base 5)4x

Answer 1

log (base5)(x-1) +log(base5) x
=log(base5)((x-1)*x)
=log(base5)(4x)
==>
(x-1)*x=4x
==>
x^2-x-4x=0
(x-3)*x=0
x=3 and/or x=0
if x=0
==>
log(base5)(0)=undefined
thus
x=3

Answer 2

first do not forget
log (base x)a + log(base x)b = log(base x)ab

so the first term is log (base5) x (x-1)

and so log (base5) x(x-1) = log (base5) 4x

you need not to calculate the number after log , knowing that if the log are the same, the number are the same

so x(x-1) =4x

x^2-x = 4x

x^2-5x =0

x(x-5) =0 2 roots x=0 and x=5

x=0 must be discarded since you have a negative number with x-1
so only x=5 is good

check correct since the log are 0.861 for4 , 1.861 for 20

Answer 3

REMEMBER 2 PROPERTIES OF LOG
1. log x + log y = log xy (same base)
2. log x - log y = log x/y (same base)
now,
log (base5)(x-1)+log(base5) x=log(base5)4x
by using 1 property of log we get as
log(base5)[(x-1)x]=log(base5)4x
=>[x-1] x=4x
=> x*2-x=4x
=> x*2-x-4x=0
=> x*2-5x=0
=> x(x-5)=0
=>x=5,0
x=0(we cannot take this value since log x=0 is not a real value)
therefore x=5 (answer)

Answer 4

since they have the same base (5) according to one of the properties it equals log (base 5) (x-1) times x=log (base 5) 4x
then you can cancel the log (base 5) on both sides and you get x(x-1)=4x then solve the quadratic after distributing the left and moving the 4x to the left

Answer 5

Product belongings for logs: log, base b, M + log, base b, N = log, base b, M*N Use this belongings to rewrite the left facet of your equation log base 5 (x+sixteen) + log base 5 (x+116) = 5 log base 5 [(x+sixteen)*(x+116)) = 5 Now rewrite into exponential type. the backside is 5 and the exponent is likewise 5. (x+sixteen)*(x+116)=5^5 (x+sixteen)*(x+116)=3125 X^2 + 132x + 1856=3125 X^2 + 132x -1269=0 (x+141)(x-9)=0 x=-141 or x=9 only x = 9 will artwork interior the unique equation in view which you will no longer be able to take the log of a unfavorable variety.

Answer 6

Just rememer that:

logx + logy = log(xy)

logx - logy = log(x/y)

Assume all logs to be base 5:

log(x-1)*x = log4x

x^2 - x -4x = 0

x(x-5)=0

so x=5 (x cannot equal zero since log(base 5) 0 has no real meaning)