Question

Q1.125^x+1 = 5^2x-5 Find x​.

Answer 1

Answer:

If 5

2x−1

=1/(125)

x−3

, find x.

Let us simplify the given expression,

5

2x−1

=1/(5

3

)

x−3

5

2x−1

=1/5

3x−9

5

2x−1

=5

3x+9

Now by comparing the powers, we get

2x−1=−3x+9

2x+3x=9+1

6x=10

x=10/5

=2

Answer 2

Explanation:

Given

125^x+1= 5^2x-5

To Find

We have to find x

$\sf\huge\bold{\underline{\underline{{Solution}}}} </p><p>Solution</p><p>$

$\sf = > {125}^{x + 1} = {5}^{2x - 5} \\ =>125 </p><p>x+1</p><p> =5 </p><p>2x−5$

Note that 125 is the cube of 5 so,it can also be written as :5³= 125

$\sf = > {5}^{3(x + 1)} = {5}^{2x - 5} \\ =>5 </p><p>3(x+1)</p><p> =5 </p><p>2x−5</p><p>$

Here,we see that 5 lies on both sides so,it gets automatically cancelled

➙3x+3= 2x-5

Making Variable and constant terms together

➙3x-2x=-5-3

➙x= -8

Check:-

➙3(x+1)= 2x-5

➙3(-8+1)=2(-8)-5

➙3(-7)= -16-5

➙-21= -21

Since,LHS= RHS(Verified)

Extra shots =>

If bases are same then powers gets added in multiplication and gets substracted in division.

=>aᵐ×aⁿ= a ᵐ⁺ⁿ

=>aᵐ÷ aⁿ= a ᵐ-ⁿ

If power is 0 then it would be consider as 1

=>a⁰= 1

Two unlike terms or value having same power then the values become whole to the power of:

=>aᵐ×bᵐ= (ab)ᵐ

=>aᵐ÷bᵐ= (a/b)ᵐ.