Question

Solve for x : 2^(x-5) . 5^(x-4) = 5​

Answer 1

if decimal no solution

if multiply x =5

Explanation:

If this is a decimal point

2^(x-5) . 5^(x-4) = 5

Rewriting .5 as 2 ^-1

2^(x-5) 2 ^ -1 ^(x-4) = 5

We know that a^ b^c = a^( b*c)

2^(x-5) 2 ^(-1*(x-4)) = 5

2^(x-5) 2 ^(-x+4) = 5

We know a^ b * a^ c = a^ ( b+c)

2^(x-5 +-x+4) = 5

2^(-1) = 5

This is not true so there is no solution

If it is multiply

2^(x-5) * 5 ^(x-4) = 5

Divide each side by 5

2^(x-5) * 5 ^(x-4) * 5^-1 = 5/5

We know that a^ b * a^c = a^ ( b+c)

2^(x-5) * 5 ^(x-4 -1) = 1

2^(x-5) * 5 ^(x-5) = 1

The exponents are the same, so we can multiply the bases

a^b * c*b = (ac) ^b

(2*5) ^ (x-5) = 1

10^ (x-5) = 1

We know that 1 = 10^0

10^ (x-5) = 10 ^0

The bases are the same so the exponents are the same

x-5 = 0

x=5

Answer 2

x = 5

Explanation:

Notice that there is also a base 5 on the right hand side of the equation, therefore, let's move
`5^(x-4)` to the right by dividing both sides by it. and then re-writing the right hand side as 5 to a power:

`2^(x-5)\,*\,5^(x-4)=5\\2^(x-5)=5/5^(x-4)\\2^(x-5)=5\,*\,5^(4-x)\\2^(x-5)=5^(5-x)`

Now apply log to both sides in order to lower the exponents (where the unknown resides):

`(x-5)\,log(2)=(5-x)\,log(5)`

Notice that when x = 5, this equation is true because it makes it the identity: 0 = 0

So, let's now examine what would be the solution of x is different from 5, and we can divide by (x - 5) both sides of the equation:

`log(2)=(5-x)/(x-5) \,log(5)\\log(2)=-1\,\,log(5)\\log(2)=-log(5)`

which is an absurd because log(2) is
`\\eq` from log(5)

Therefore our only solution is x=5