How do you solve #5^(2x-1) -2 =5^(x-1/2)#?

1 Answer
Ratnaker Mehta
Jun 21, 2016

# x=1/2log_5(20).#

Explanation:

Let #(2x-1)=t.#

Observe that #RHS# of the Eqn. is #5^((2x-1)/2).#

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

Next, let #5^(t/2)=y#, so that, #5^t={5^(t/2)}^2=y^2.#

Now the eqn. is, #y^2-2=y,# or, #y^2-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).#

Related questions
See all questions in Logarithm-- Inverse of an Exponential Function
Impact of this question
1484 views around the world
You can reuse this answer
Creative Commons License