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

mathsworkmusic | Certified Educator

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

`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

stacy143 | Student

`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

kspcr111 | Student

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`

vamkitten | Student

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

honeybujji | Student

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.