How can I solve log(x)+3x=5?

Question

How can I solve matematically log(x)+3x=5

Thank you in advance

Thank you very much for your answer. There are very usefull.

Answer 1

1.52582219663167... if the log is base e
1.59874065990736... if the log is base 10

As the use of a graphing calculator does not give any insight into the derivation of the solution, here it is:

To solve this, you need to use some numerical interpolation method, as the equation can not be solved directly for x (in terms of finitely many algebraic functions).

Newton's method yields:
Let y = f(x) = log(x) + 3x - 5
{here log(x) is the natural log(base e). We want to find the value of x for which f(x) = 0}
f '(x) = 1/x + 3
the method says that x(n+1) = x(n) + (y - f(x(n)) / f '(x(n))
and here, y = 0, so

x(n+1) = x(n) + (5 - log(x(n)) - 3x(n)) / (1/x(n) + 3)

then, x = lim(n→∞) x(n)

(You must choose something reasonably close for x(1), such as
x(1) = 1. It takes only five iterations to get the accuracy of my solutions above in a VB6 program using double precision variables. i.e., x(5)=1.52582219663167)

Answer 2

i will clarify the 1st occasion, and only clean up the rest. First, you are able to convert each and every thing to log. as an occasion, 2 is log2(4). So the equation decrease than a) could appear like: log2(4)+log2(x)=2log2(x+a million) Then we are able to merge the 1st 2, because of the fact the sum of two logs is the log of the made of their parameters. occasion: log2(4)+log2(x)=log2(4*x). ok, so now our equation in log2(4*x)=2log2(x+a million). we are able to additionally positioned the two after the equals sign into the logarithm, as an exponent. 2log2(x+a million)=log2(x+a million)^2 which makes the equation log2(4x)=log2(x+a million)^2. we are able to now actual do away with the logarithms, and are left with in basic terms 4x=(x+a million)^2. 4x=x^2+2x+a million x^2-2x+a million=0 x=a million. b) log(x) + log(3x+a million) = a million log(x*(3x+a million)) = log(10) 3x^2+x=10 3x^2+x-10=0 x1=5/3, x2=-2 yet x>a million, so we in basic terms take the 1st answer x=5/3. For the final 2, we would desire to continuously use the undeniable fact that loga(a^x) = x And for the log2(4000) and log3(250) I only used a calculator, i don't comprehend the way you're envisioned to do it, yet there is not any incorrect way for those numbers :/ c) 2^x = 4000 log2(2^x) = log2(4000) x = 11.966 d) 3^(y+2) = 250 log3(3^(y+2)) = log3(250) y+2 = 5.026 y = 3.026

Answer 3

There are complex mathematical ways to calculate this, but from a practical perspective, you can quickly use MS-Excel:

It just takes 2 cells, the first is for a value for x, the second is the equation i.e. if cell C3 is used for x, fill a cell with '=log(C3)+3*C3.

You can trial by error play with the value for x until your result=5, but Excel has a useful little tool, under 'Tools' choose 'Goal Seek' use that excel will find the value of x for you.

Answer 4

That's a 'transcendental equation' and cannot be solved directly. It requires a numbeical solution. Both log x and 3x get bigger with bigger x. Plus, log 1 = 0, so you know that there is a solution. On my TI-83:

Graph both log x and 5-3x They cross at about x = 1.6

log(1.6)= 0.204
5-3(1.6) = 0.2000

That's close.

Note: I've assumed that you want base 10 log.

Using base e log, and the intersection function of my TI-83 I get
x = 1.526

Answer 5

WEll i did not really get an exact answer, but i did get what x= to but its with a variable in it, so here it goes, maybe it helps you or something clicks and you get it.

log(x)+3x=5
log(x)=5 - 3x
10^(5-3x) = x

I do not know what to do from there if it says solve for x, but here you go. thats all i could come up with, cant think further.