# Please solve  `2^(x+1)*5^(2x-1)=40*0.1^(x+2).` I don't know how to solve these equations when the numbers multiply. Please explain!

Luca B. | Certified Educator

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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.`

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