# lg 4x = 2 - 2lg5lg 4x = 2 - 2lg5

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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:

1

**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).

We get

lg 4x + 2*lg 5=2

=> lg 4x + lg 5^2=2

=> lg 4x + lg 25=2

=> lg 100x=2

=> 100x= e^2

=>x= e^2/100

**Therefore x= (e^2)/100.**

lg4x= 2-2lg5.

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.

Take antilog:

4x = 4

x = 4/4 = 1

x = 1.

Before solving the equation, we'll have to impose constraints of existence of logarithms.

4x>0

We'll divide by 4:

x>0

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.