Respuesta :
The equation is
[tex]\log_{64}(25x)-\log_{64}(x-7)=\frac{2}{3}[/tex].
First, from the rule [tex]\log_ba-\log_bc=\log_b{\frac{a}{c}}[/tex], we rewrite the left hand side as:
[tex]\displaystyle{ \log_{64}{ \frac{25x}{x-7} }[/tex].
Now we write the right hand side as a log with base 64, so that we can make the arguments equal, setting a new equation.
[tex]\displaystyle{ \frac{2}{3}= (\frac{2}{3})\log_{64}64=\log_{64}64^{\frac{2}{3}}[/tex].
We can simplify the argument of this expression into a more manageable expression using the laws of exponents:
[tex]64^{\frac{2}{3}}=(2^6)^{\frac{2}{3}}=2^{6\cdot\frac{2}{3}}=2^4=16[/tex].
Thus, from [tex]\displaystyle{ \log_{64}({ \frac{25x}{x-7} })=\log_{64}(16)[/tex] we clearly have:
[tex]\displaystyle{ \frac{25x}{x-7}=16[/tex].
For x different from 7:
25x=16(x-7)
25x=16x-112
9x=-112
x=-112/9.
This value is approximately -12.44, so x-7, the argument of both the first and the second expression, would be negative. Thus, the equation has no solution.
[tex]\log_{64}(25x)-\log_{64}(x-7)=\frac{2}{3}[/tex].
First, from the rule [tex]\log_ba-\log_bc=\log_b{\frac{a}{c}}[/tex], we rewrite the left hand side as:
[tex]\displaystyle{ \log_{64}{ \frac{25x}{x-7} }[/tex].
Now we write the right hand side as a log with base 64, so that we can make the arguments equal, setting a new equation.
[tex]\displaystyle{ \frac{2}{3}= (\frac{2}{3})\log_{64}64=\log_{64}64^{\frac{2}{3}}[/tex].
We can simplify the argument of this expression into a more manageable expression using the laws of exponents:
[tex]64^{\frac{2}{3}}=(2^6)^{\frac{2}{3}}=2^{6\cdot\frac{2}{3}}=2^4=16[/tex].
Thus, from [tex]\displaystyle{ \log_{64}({ \frac{25x}{x-7} })=\log_{64}(16)[/tex] we clearly have:
[tex]\displaystyle{ \frac{25x}{x-7}=16[/tex].
For x different from 7:
25x=16(x-7)
25x=16x-112
9x=-112
x=-112/9.
This value is approximately -12.44, so x-7, the argument of both the first and the second expression, would be negative. Thus, the equation has no solution.
