The problem involves solving the logarithmic equation 9 log3(2x+1) + 4 log2(x+3) = 85, which includes logarithms with different bases. Participants are exploring methods to approach the problem, particularly focusing on the implications of the change of base rule.
The discussion is ongoing, with various participants offering insights into numerical methods for solving the equation. Some have suggested approximations and numerical techniques, while others are still grappling with the algebraic manipulation required.
There is mention of the difficulty in obtaining a closed form solution, with some participants indicating that numerical methods may be necessary. The original poster has not received explicit guidance from the teacher on how to solve the problem, leading to a reliance on peer suggestions.
jgens said:Do you know about the change of base rule?
Yes, that's right. And that changes the equation toharimakenji said:I am not really sure what you meant. If this what you meant log3(2x+1) = log2(2x+1) / log23, I have tried that and do not know how to continue
Thank you very much
HallsofIvy said:Yes, that's right. And that changes the equation to
\frac{9}{log_2(3)}log_2(2x+1)+ 4log_2(x+ 3)= 85
Now, as JonF suggested, use the laws of logarithms.
\frac{9}{log_2(3)}log_2(2x+1)= log_2((2x+1)^{9/log_2(3)}
and
4 log_2(x+3)= log_2((x+3)^4)
so we have
log_2((2x+1)^{9/log_2(3)}(x+3)^4}= 85
and you can take 2 to the power of each side to get
(2x+1)^{9/log_2(3)}(x+3)^4= 2^{85}
Now, of course, that's still a pretty nasty equation! As jgens said, you probably will need to solve this numerically.
JonF said:Through incredibly awful calculus or punching it in a calculator. If some horrible teacher forced me to find a numerical approximation of x, I would first bring everything to the RHS, then call the LHS f(x). And do some numerical approximation of zeros method like Newtons or graph it very carefully and eyeball it.
BobG said:Or approximate it, assuming x was close to some power of 2, then brute force it with excel. Pull the 2 up into the power on the first term and...
That would give you approximately:
x^{6.678}*x^4=2^{85}
x^{10.678}=2^{85}
x=2^8
x is a little more than 256, which at least gives you a decent starting point if you're using Newton's method or a decent starting point if you decide to just brute force it. You can get to 291 (actually, closer to 292) very quickly just by brute forcing it, which is already more significant digits than you started with.