You should use the exponential identities such that:

`2^(x+1) = 2^x*2`

`5^(2x-1) = 5^(2x)*(5^(-1)) => 5^(2x-1) = 25^x*(1/5)`

Performing the multiplication yields:

`2^(x+1)*5^(2x-1) = 2^x*2*25^x*(1/5)`

Using the identity `(a^x*b^x)= (a*b)^x ` yields:

`2^x*25^x = (2*25)^x = 50^x`

`2^(x+1)*5^(2x-1) = (2/5)*50^x`

`0.1^(x+2) = 0.1^x*0.1^2 => 1/10^x*1/100`

`40*0.1^(x+2) = 4/10*1/10^x`

`(2/5)*50^x = 4/10*1/10^x => 50^x = 10^(-x)`

You need to take logarithms both sides such that:

`log (50^x) = log (10^(-x))`

Using logarithmic identities yields:

`x*log 50 = -x*log 10`

Since `log 10 = 1` yields:

`x*log 50 = -x => x*log 50 + x= 0`

You need to factor out x such that:

`x(log 50 + 1) = 0`

Since `log 50 + 1 != 0 => x = 0.`

**Hence, evaluating the solution to the given equation yields `x = 0.` **

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now