Question #3624b

`5^(2x-3)-2^(5x+2)=0`
`5^(2x-3)=2^(5x+2)`
Taking the natural logarithm of both sides,
`ln(5^(2x-3))=ln(2^(5x+2))`
`(2x-3)ln5=(5x+2)ln2`
`(2ln5)x-3ln5=(5ln2)x+2ln2`
`(2ln5)x-(5ln2)x=3ln5+2ln2`
`[ln(5^2)-ln(2^5)]x=ln(5^3)+ln(2^2)`
`x=ln(5^3*2^2)/ln(5^2/2^5)`
`x=ln(125*4)/ln(25/32)`
`x=log_(25/32)500`

Given: `5^(2x-3)-2^(5x+2)= 0`

Move the second term to the right:

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

Use the base 5 logarithm on both sides:

`log_5(5^(2x-3))=log_5(2^(5x+2))`

Use the identity `log_b(a^c) = (c)log_b(a)` on both sides:

`(2x-3)log_5(5)=(5x+2)log_5(2)`

Use the property `log_b(b) = 1` on the left:

`2x-3=(5x+2)log_5(2)`

Use the distributive property on the right:

`2x-3=5log_5(2)x+2log_5(2)`

Subtract `5log_5(2)x` from both sides:

`(2-5log_5(2))x-3=2log_5(2)`

Add 3 to both sides:

`(2-5log_5(2))x=3+2log_5(2)`

Divide both sides by the coefficient of x:

`x=(3+2log_5(2))/(2-5log_5(2))`

Convert to base e by using the conversion formula `log_5(x)= ln(x)/ln(5)`:

`x=(3+2ln(2)/ln(5))/(2-5ln(2)/ln(5))`

Multiply by 1 in the form of `ln(5)/ln(5)`:

`x=(3ln(5)+2ln(2))/(2ln(5)-5ln(2))`