How do you solve 5^(2x-1) -2 =5^(x-1/2)52x−1−2=5x−12?

1 Answer
Ratnaker Mehta
Jun 21, 2016

x=1/2log_5(20).x=12log5(20).

Explanation:

Let (2x-1)=t.(2x−1)=t.

Observe that RHSRHS of the Eqn. is 5^((2x-1)/2).52x−12.

Hence, the given eqn. becomes, 5^t-2=5^(t/2)5t−2=5t2.

Next, let 5^(t/2)=y5t2=y, so that, 5^t={5^(t/2)}^2=y^2.5t={5t2}2=y2.

Now the eqn. is, y^2-2=y,y2−2=y, or, y^2-y-2=0.y2−y−2=0.

:. y^2-2y+y-2=0.
:. y(y-2)+1(y-2)=0.
:. (y-2)(y+1)=0.
:. y=2, y=-1, but, y being 5^(t/2) can not be -ve, so, y!=-1.
:. y=2 rArr 5^(t/2)=2 rArr 5^t={5^(t/2)}^2=2^2=4.
:. t=log_5(4)
:. 2x-1=log_5(4).
:. 2x=1+log_5(4),=log_5(5)+log_5(4)=log_5(20).
:. x=1/2log_5(20).

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