Respuesta :

[tex]5^{2x}=7^{x-1}\\ \ln5^{2x}=\ln7^{x-1}\\ 2x \ln 5 =(x-1)\ln 7\\ 2x \ln 5 =x\ln 7-\ln 7\\ 2x\ln 5-x\ln 7=-\ln 7\\ x(2\ln5 -\ln 7)=-\ln7 \\ x=-\dfrac{\ln 7}{2\ln 5-\ln 7}\\ x\approx -1.52864 [/tex]

Answer:

option C is correct.

Value of x = -1.52864

Step-by-step explanation:

Solve: [tex]5^{2x} = 7^{x-1}[/tex]                  .....[1]

Using logarithmic rule:

[tex]\ln a^n = n \ln a[/tex]

Taking log with base e both sides in [1] we get;

[tex]\ln 5^{2x} = \ln 7^{x-1}[/tex]

Apply logarithmic rules; we have

[tex]2x \ln 5 = (x-1) \ln 7[/tex]

Using distributive property: [tex]a \cdot ( b+c) = a\cdot b + a\cdot c[/tex]

[tex]2x \ln 5 = x\ln 7-\ln 7[/tex]

Add both sides [tex]\ln 7[/tex] we get;

[tex]2x \ln 5 +\ln 7= x\ln 7[/tex]

or

[tex]\ln 7= x\ln 7-2x \ln 5[/tex]

[tex]\ln 7= x(\ln 7-2\ln 5)[/tex]

or

[tex]x = \frac{\ln 7}{\ln 7- 2\ln 5}[/tex]

Simplify:

x = -1.528643062439 ≈ - 1.52864

Therefore, value of x is, -1.52864