logarithmic problem still struggling

• Feb 1st 2007, 11:01 AM
kcsteven
logarithmic problem still struggling
I have worked this out different ways and I am not comming up with the correct answer, although I have gotten many answers. Here it is:
solve for x
9/7log3x=8
(The 3 is the base but I don't know how to type it so it is lower than the rest of the numbers)
I know I am supposed to solve for log3x first and then use the fact that if log ax=b, then x=a^b but the answers don't jive with the book. The first one I tried I got 9/7log3^8=x then I came up with 72/7log3, the three is the base.
I know this is not correct, could you explain so I can understand.
Thank you!
Keith Stevens
ps. I just want to say a big thank you to everyone that helps on this site because when we are not at school or the teacher or our friends are too busy to help, this site is an excellent place to get the information needed to complete the rest of our homework or to be able to grasp the concepts and move ahead. Anyway thankx a million everyone!!!!!!!!
Keith Stevens
• Feb 1st 2007, 11:24 AM
topsquark
Quote:

Originally Posted by kcsteven
I have worked this out different ways and I am not comming up with the correct answer, although I have gotten many answers. Here it is:
solve for x
9/7log3x=8
(The 3 is the base but I don't know how to type it so it is lower than the rest of the numbers)
I know I am supposed to solve for log3x first and then use the fact that if log ax=b, then x=a^b but the answers don't jive with the book. The first one I tried I got 9/7log3^8=x then I came up with 72/7log3, the three is the base.
I know this is not correct, could you explain so I can understand.
Thank you!
Keith Stevens
ps. I just want to say a big thank you to everyone that helps on this site because when we are not at school or the teacher or our friends are too busy to help, this site is an excellent place to get the information needed to complete the rest of our homework or to be able to grasp the concepts and move ahead. Anyway thankx a million everyone!!!!!!!!
Keith Stevens

Looks like a simple Math mistake to me. (Of the type that's almost impossible to catch once you've done it.)

Solve:
$\frac{9}{7}log_3(x) = 8$

$log_3(x) = \frac{56}{9}$

Thus:
$x = 3^{56/9}$

-Dan
• Feb 1st 2007, 11:33 AM
earboth
Quote:

Originally Posted by kcsteven
I have worked this out different ways and I am not comming up with the correct answer, although I have gotten many answers. Here it is:
solve for x
9/7log3x=8
(The 3 is the base but I don't know how to type it so it is lower than the rest of the numbers)
...
Keith Stevens

Hello, Keith,

unfortunately you didn't word this problem unambiguously:

1. possibility:

$\frac{9}{7} \log_{3}{x}=8 \Longleftrightarrow \log_{3}{x}= \frac{7 \cdot 8}{9}$

You know that $b^{\log_{b}{x}}=x$. Therefore use the base 3 on both sides of this equation:

$3^{\log_{3}{x}}=3^{\frac{7 \cdot 8}{9}} \approx 930.58...$

2. possibility:

$\frac{9}{7\cdot \log_{3}{x}}=8 \Longleftrightarrow \log_{3}{x}= \frac{9}{7 \cdot 8}$

Use the base 3 on both sides of this equation:

$3^{\log_{3}{x}}=3^{\frac{9}{7 \cdot 8}} \approx 1.1931...$

EB