3^(4x-5) = (1/27)^(2x+10) show your work

3^4x-5=1/27^2x+10

We move all terms to the left:

3^4x-5-(1/27^2x+10)=0

Domain of the equation: 27^2x+10)!=0

x∈R

We get rid of parentheses

3^4x-1/27^2x-10-5=0

We multiply all the terms by the denominator

3^4x*27^2x-10*27^2x-5*27^2x-1=0

Wy multiply elements

81x^6-270x^2-135x^2-1=0

We do not support expression: x^6    

Step-by-step explanation:

step-by-step explanation: h: (- 3, 6)

m: (1, - 2)

o: (7, 6)

r: (- 9, - 2)

midpoint of hm: (- 1, 2)

midpoint of or: (- 1, 2)

therefore, b is the midpoint of both diagonals bisect each other !

to prove the diagonals are perpendicular, show that the product of the slopes = - 1

the answer is 3/2 and for decimal: 1.5

hope i

-lavira: )