196^x - 4*14^x + 3 = 0.

We know that 196 = 14^2:

Let us rewrite:

(14^x)^2 - 4*14^x + 3 = 0

Let us asume that y= 14^x

==> y^2 - 4y + 3 = 0

Let us factor:

==> (y-3)(y-1) = 0

Then we have two possible solution:

y1= 3 ==> 3= 14^x1 ==> **x1= log 14 (3)**

y2= 1 ==> 1= 14^x2 ==> **x2= 0**

To solve this equation lets create a quadratic equation first:

Equate 14^x=y, this gives 196^x= 14^2^x= (14^x)^2 = y^2

So we have: y^2- 4*y+3=0

=> y^2- 3y – y + 3 = 0

=> y* (y-3) – 1* (y-3) = 0

=> (y-1) (y-3) = 0

Therefore y can be 1 and 3

Now y=14^x

Now 14^x can be 3 when

log (14^x)= log 3

=> x log 14 = log 3

=> x = (log 3 / log14).

14^x = 1 for x= 0.

**Therefore x can be 0 and log3 / log14.**

we note that:

196 = 14^2

Therefore the given equation can be rewritten as

14^2x - 4*14^x + 3 = 0

Let:

14^x = a

Then substituting a in place of 14^x the given equation becomes:

a^2 - 4a + 3 = 0

a^2 - 4a + 4 - 1 = 0

(a - 2)^2 - 1 = 0

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

(a - 1)(a - 3) = 0

Therefore:

a = 1 and

a = 3

Therefore:

14^x = 1

and

14^x = 3

When 14^x = 1 then:

x = 0

And when 14^x = 3

x = 0.4162866 (approximately)

Thus x has two possible values:

x = 0, 0.4162866

196^x-4*14^x+3 = 0

To solve for x.

196^x = (14^x). So we put 14^x = y and the equation becomes:

y^2-4y+3 = 0

(y-1)(y-3) = 0

So y-1 = 0 gives 14^x = 1. Or 14^x = 14^0. So x = 0.

y-3 = 0 gives 14^x = 3. Or

x log 14 = log3.

So x= log3/log14.

The equation is an exponential equation. We'll re-write the equation substituting 196^x by 14^2x.

14^2x - 4*14^x + 3 = 0

It is advisable to solve this type of equation, using substitution method.

Now we can do the substitution 14^x=t

t^2 - 4*t + 3 = 0

This is a quadratic equation and to find it's roots, we can apply the quadratic formula:

t1 = [4+sqrt(16-12)]/2

t1 = (4+2)/2

t1 = 3

t2 = (4-2)/2

t2 = 1

But the initial equation is not solved yet.

14^x = 3

We'll take logarithms both sides:

log 14 (14^x) = log 14 3

x*log 14 14 = log 14 3

x = log 14 3

14^x = 1

14^x = 14^0

x = 0

**The equation has 2 real solutions: {0 , log 14 3}.**