Logarithm question

erice1977

New member
Joined
Nov 25, 2009
Messages
2
I am having trouble figuring out how to start this problem.

solve for x: (5^x - 5^-x)/(5^x + 5^-x) = 8

Do I use log to both sides giving this--> log(5^x - 5^-x) - log(5^x + 5^-x) = log8 ? if i do this what is the next step?
or
divide the denominator? or do i need to make the exponents positive? any help would appreciated
 
Hello, erice1977!

There may be a typo . . . as written, the equation has no real solutions.
A sign-change will produce a real answer.


\(\displaystyle \text{Solve for }x\!:\;\;\frac{5^x + 5^{-x}}{5^x - 5^{-x}} \:=\:8\)

\(\displaystyle \text{We have: }\;5^x + 5^{-x} \:=\:8\!\cdot\!5^x - 8\!\cdot\!5^{-x}\)

. . . . . . . \(\displaystyle 7\!\cdot\!5^x \;=\;9\!\cdot\!5^{-x}\)

\(\displaystyle \text{Multiply by }5^x\!:\;\;7\cdot5^{2x} \:=\:9 \quad\Rightarrow\quad 5^{2x} \:=\:\tfrac{9}{7}\)

\(\displaystyle \text{Take logs: }\;\ln\left(5^{2x}\right) \;=\;\ln\left(\tfrac{9}{7}\right) \quad\Rightarrow\quad 2x\cdot\ln(5) \;=\;\ln\left(\tfrac{9}{7}\right)\)

\(\displaystyle \text{Therefore: }\;x \;=\;\frac{\ln\left(\frac{9}{7})}{2\ln(5)}\)

 
Thank you for taking time to respond. However, the problem is written correctly. If there are no real solutions, that is acceptable, but I need to show my work.

In the answer you provided, where did the 9 and the 7 come from?, sorry I am confused maybe break it into smaller steps?

Thanks again.
 
erice1977 said:
Thank you for taking time to respond. However, the problem is written correctly. If there are no real solutions, that is acceptable, but I need to show my work.

In the answer you provided, where did the 9 and the 7 come from?, sorry I am confused maybe break it into smaller steps?

Thanks again.
\(\displaystyle 5^x + 5^{-x} = 8 * 5^x - 8*5^{-x}\)

\(\displaystyle 8 * 5^{-x} + 5^{-x} = 8 * 5^x - 5^{x}\)

Instead of staring at the screen, take paper and pencil and go step by step - it should be very clear.

For your "actual" problem - follow the same steps and derive the conclusion suggested by Soroban.
 
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