# How would you solve a problem like this? Solve the equation `25^(2x+3)= 125^ (2x+8)` (please explain the steps in detail)

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**Solve the equation for** `x ` :

`25^(2x+3) = 125^(2x+8) `

First notice that we can simplify by realising that 25 is 5 squared and 125 is 5 cubed so that we have

`5^(2(2x+3)) = 5^(3(2x+8)) ` implying (by multiplying out the brackets)

`5^(4x+6) = 5^(6x + 24) `

* At this stage, we could simply equate the power terms, since the base on each side (5) is the same, and solve for x*. However, if the bases weren't the same, the standard approach would be to take the logarithm of each side ('take logs'). So let's look at this. We can write that

`log[5^(4x+6)] = log[5^(6x+24)] `

When we 'take logs' the base is typically 10 or `e ` (the latter corresponding to the 'natural logarithm'). However, we can in fact choose any base we like. Now, since * the logarithm of a number is the power to which the base needs to be raised to obtain that number*, it is plain here that choosing

*would be sensible. Taking the base of the logarithmic terms in the equation as 5, the terms on either side simplify to:*

**base 5**`log_5[5^(4x+6)] = 4x +6 ` since `5 ` must be raised to the power `4x+6 ` to obtain `5^(4x+6) `

and similarly

`log_5[5^(6x+24)] = 6x + 24 `

Equating these two sides of the equation we have that

`4x +6 = 6x + 24 `

Finally, gather terms in `x ` on the left-hand side of the equation and constant terms on the right-hand side, giving

`-2x = 18 ` that is

` x = -9` **answer**

`25^(2x+3)= 125^(2x+8) `

Applying log on both sides , we get :

`log(25^(2x+3))= log(125^(2x+8))`

= ` (2x+3)*log(25)= (2x+8)*log(125) `

= `(2x+3)*log(5^2)= (2x+8)*log(5^3)`

= `2*(2x+3)*log(5)= 3*(2x+8)*log(5)`

= ` 2*(2x+3)= 3*(2x+8)`

= 4x+6 =6x+24

= 2x= -18

x=-9

Given `25^(2x+3)=125^(2x+8)`

now first we need to make the bases equal , proceeding as follows

`5^(2(2x+3))=5^(3(2x+8))`

`5^(4x+6)=5^(6x+24)`

As the bases are now equal, so the powers can be equated as follows

`=> 4x+6 = 6x+24`

` => 4x-6x=24-6`

`=> -2x =18`

` => x = -9`

First 25^2x+3=5^2(2x+3)

second 125^2x+8=5^3(2x+8)

third cancelled the 5 and left the powers so it will be

2(2x+3)=3(2x+8)

4x+6=6x+24

-2x=18

-x=9

x=-9

25 2x+3 = 125 2x=8

First express 25 and 125 in terms of power that will make 25 = 5 to the power of 2 and 125 = 5 to the power of 3 when both the bases are equal then we can solve the sum as

LHS = 2(2x+3) = 4x +12

RHS = 3(2x+8) = 6x+24 Now equate both LHS and RHS which becomes

4x+12 = 6x+24 ; bring the variable to the same side

4x-6x = 24-12 = -2x = 12 then - x will be 12/-2 = -6 so when we cancel both the minus signs on both sides then x will be 6.