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