Respuesta :

mberisso

Answer:

x = 5

Step-by-step explanation:

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

[tex]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}[/tex]

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

[tex](x-5)\,log(2)=(5-x)\,log(5)[/tex]

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:

[tex]log(2)=\frac{5-x}{x-5} \,log(5)\\log(2)=-1\,\,log(5)\\log(2)=-log(5)[/tex]

which is an absurd because log(2) is [tex]\neq[/tex] from log(5)

Therefore our only solution is x=5

wegnerkolmp2741o

Answer:

if decimal  no solution

if multiply x =5

Step-by-step 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

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