how to solve for n

[mimie]

[MATH]6561=(-3)^{(n-1)} [/MATH]I try to convert to log form, but the base is negative, and calculator give me math error.
Please teach me how to solve this question, thanks.

 

[Deleted member 4993]

[MATH]6561=(-3)^{(n-1)} [/MATH]I try to convert to log form, but the base is negative, and calculator give me math error.
Please teach me how to solve this question, thanks.

Can you solve for 'n' when:

6561 = 3^(n-1)

 

[mimie]

Can you solve for 'n' when:

6561 = 3^(n-1)

what happen to the negative sign? how to make negative sign disappear?

 

[pka]

[MATH]6561=(-3)^{(n-1)} [/MATH]I try to convert to log form, but the base is negative, and calculator give me math error.

Mimie, IF there is a solution, can you explain why \(\displaystyle (n-1)\) must be even? LOOK HERE

 

[mimie]

Mimie, IF there is a solution, can you explain why \(\displaystyle (n-1)\) must be even? LOOK HERE

[MATH] \displaystyle 6561 = (-3)^{(n-1)} \\= (-1 \times 3)^{(n-1)} \\= (-1)^{(n-1)} \times3^{(n-1)} \\6561 >0 \\ 3^{(n-1)}>0 \\(-1)^{(n-1)}>0 \\so\quad(-1)^{(n-1)}=1 \quad and\quad\:n-1 \: is\: even\: integer \\6561 = (3)^{(n-1)} \\ \log _{3} 6561=n-1 \\8=n-1 \\n=9 [/MATH]
please check my calculation, is it correct? Thanks.

 

[pka]

[MATH] \displaystyle 6561 = (-3)^{(n-1)} \\= (-1 \times 3)^{(n-1)} \\= (-1)^{(n-1)} \times3^{(n-1)} \\6561 >0 \\ 3^{(n-1)}>0 \\(-1)^{(n-1)}>0 \\so\quad(-1)^{(n-1)}=1 \quad and\quad\:n-1 \: is\: even\: integer \\6561 = (3)^{(n-1)} \\ \log _{3} 6561=n-1 \\8=n-1 \\n=9 [/MATH]
please check my calculation, is it correct? Thanks.

Correct, well done.

 

[HallsofIvy]

If n is odd then n-1 is even and (-3)^(n-1)= 3^(n-1). If n is even then n-1 is odd and (-3)^(n-1)= -3^(n-1). In any case, no "logarithms" are necessary because 6561= 3^8. The problem is to find n such that (-3)^(n-1)= 3^8. Yes, "8" is even so we can ignore the negative. We have n-1= 8 so n= 9.

 

[Steven G]

[MATH] \displaystyle 6561 = (-3)^{(n-1)} \\= (-1 \times 3)^{(n-1)} \\= (-1)^{(n-1)} \times3^{(n-1)} \\6561 >0 \\ 3^{(n-1)}>0 \\(-1)^{(n-1)}>0 \\so\quad(-1)^{(n-1)}=1 \quad and\quad\:n-1 \: is\: even\: integer \\6561 = (3)^{(n-1)} \\ \log _{3} 6561=n-1 \\8=n-1 \\n=9 [/MATH]
please check my calculation, is it correct? Thanks.

I too am impressed with your work! As far as us checking your work you can do that yourself. Just verify if 3^8 = 6561 or even better, check if (-3)^8 =6561