lg 4x = 2 - 2lg5lg 4x = 2 - 2lg5
You need to write the constant term 2 as since such that:
You need to factor out 2 such that:
You need to convert the difference of logarithms into a quotient such that:
Hence, evaluating the solution to the given equation yields
lg 4x = 2 - 2lg5
Move 2lgx to the left side:
==> lg 4x + 2lg 5= 2
We know that: a*log b = log b^a
==> lg 4x + lg 5^2 = 2
==> lg 4x + lg 25 = 2
We know that log a + log b = log a*b
==> lg 4x*25 = 2
==> lg 100x = 2
==> 100x = 10^2
==> 100x = 100
==> x= 1
We have to find the value of x that meets the equation lg 4x= 2- 2*lg 5.
lg 4x= 2- 2*lg 5
=> lg 4x + 2*lg 5=2
Now we use the property of logarithms that a*lg b= lg (b^a) and lg a+ lg b= lg (a*b).
lg 4x + 2*lg 5=2
=> lg 4x + lg 5^2=2
=> lg 4x + lg 25=2
=> lg 100x=2
=> 100x= e^2
Therefore x= (e^2)/100.
To solve for x.
By property of logarithms, mlog a = lg a^m and lg a - lg b = loga/b. 2 = lg100 . We now rewrite the equation:
l4x = lg100 - lg5^2
lg4x = lg (100/25) = lg 4
lg4x = lg 4.
4x = 4
x = 4/4 = 1
x = 1.
Before solving the equation, we'll have to impose constraints of existence of logarithms.
We'll divide by 4:
Now, we could apply the power property of logarithms:
2log 5 = log 5^2 = log 25
We'll re-write the equation:
log 4x + log 25 = 2
Now, we'll apply the product property of logarithms:
log 4x + log 25 = log 4*25*x = log 100x
log 100x = 2 => 100x = 10^2 => 100x = 100
We'll divide by 100 both sides:
x = 1 >0
Since the value of x is in the interval (0, +inf.), the solution is admissible.