Logarithms: x^2 log10^8 - x log10^5 = 2(log2^10)^-1-x

[cyberspace]

Solve for x:
x^2 log10^8 - x log10^5 = 2(log2^10)^-1-x

It took me one hour to do this, and I got : x(x log10^8/5) = 1/log2^100-x, and I'm not sure if this is right.
How do I get x?

I apologize if I'm being a pest!

 

[stapel]

Solve for x: x^2 log10^8 - x log10^5 = 2(log2^10)^-1-x

This is just a quadratic with some extremely "lumpy" coefficients, before simplification:

. . . . .(log(10[sup:ujh6ghir]8[/sup:ujh6ghir]))x[sup:ujh6ghir]2[/sup:ujh6ghir] - (log(10[sup:ujh6ghir]5[/sup:ujh6ghir]))x = 2(log(2[sup:ujh6ghir]10[/sup:ujh6ghir])[sup:ujh6ghir]-1[/sup:ujh6ghir]) - x

. . . . .(log(10[sup:ujh6ghir]8[/sup:ujh6ghir]))x[sup:ujh6ghir]2[/sup:ujh6ghir] - (log(10[sup:ujh6ghir]5[/sup:ujh6ghir]))x + x - 2(log(2[sup:ujh6ghir]10[/sup:ujh6ghir])[sup:ujh6ghir]-1[/sup:ujh6ghir]) = 0

Since the logs are (presumably) base 10, then the log expressions simplify to give:

. . . . .8x[sup:ujh6ghir]2[/sup:ujh6ghir] - 5x + 1x - 1/(5 log(2)) = 0

. . . . .8x[sup:ujh6ghir]2[/sup:ujh6ghir] - 4x - 1/(5 log(2)) = 0

Now apply the Quadratic Formula. :wink:

It took me one hour to do this, and I got : x(x log10^8/5) = 1/log2^100-x,....

It is unfortunate that you did not provide your steps. I'm afraid it is difficult to comment on work we can't see. Sorry!

Eliz.

 

[cyberspace]

Sorry for the confusion,

In the equation when I wrote log 10^8 and many more, I meant that "10" is the base and "8" is the argument.

 

[stapel]

In the equation when I wrote log 10^8 and many more, I meant that "10" is the base and "8" is the argument.

Oh. That changes things... :shock:

So you actually meant the following?

. . . . .x[sup:25lcjgur]2[/sup:25lcjgur] log[sub:25lcjgur]10[/sub:25lcjgur](8) - x log[sub:25lcjgur]10[/sub:25lcjgur](5) = 2 (log[sub:25lcjgur]2[/sub:25lcjgur](10))[sup:25lcjgur]-1[/sup:25lcjgur] - x

What were your steps?

Note: By the change-of-base formuls, you can convert log[sub:25lcjgur]2[/sub:25lcjgur](10) to log[sub:25lcjgur]10[/sub:25lcjgur](10)/log[sub:25lcjgur]10[/sub:25lcjgur](2) = 1/log[sub:25lcjgur]10[/sub:25lcjgur](2). Since this log was in the denominator (due to the negative power), you would get:

. . . . .x[sup:25lcjgur]2[/sup:25lcjgur] log[sub:25lcjgur]10[/sub:25lcjgur](8) - x log[sub:25lcjgur]10[/sub:25lcjgur](5) = 2 log[sub:25lcjgur]10[/sub:25lcjgur](2) - x

...or:

. . . . .x[sup:25lcjgur]2[/sup:25lcjgur] log[sub:25lcjgur]10[/sub:25lcjgur](8) - x (log[sub:25lcjgur]10[/sub:25lcjgur](5) + 1) - 2 log[sub:25lcjgur]10[/sub:25lcjgur](2)= 0

Then apply the Quadratic Formula. You may (or may not; I haven't checked) find it helpful to note that 1 = log[sub:25lcjgur]10[/sub:25lcjgur](10), so log[sub:25lcjgur]10[/sub:25lcjgur](5) + 1 = log[sub:25lcjgur]10[/sub:25lcjgur](50).

Eliz.