Multiple in logarithm question -- find the variable?

[Helly123]

Homework Statement

Homework Equations

$$ \log_{x} a \ means \ 10^x = a $$
$$ \log_{x} {a^n} = n \log_{x}{a} $$

The Attempt at a Solution

$$ \log_{10} {ax} \log_{10}{bx} + 1 = 0 $$
$$ \ means \log_{10} {ax} = 1 \log_{10}{bx} = -1 \ or \ vice \ versa $$
$$ \ for \log_{10} {ax} = 1 \ so \ ax \ = 10 \ and \ \log_{10}{bx} = -1 \ so \ bx \ = 0.1 \ or \ vice \ versa $$
bx/ax = b/a = 0.1/10 = 0.01
or b/a = 10/0.1 = 100
so b/a > ?
how to make range for b/a ? how's the correlation with x?

 

[Mark44]

Homework Statement

View attachment 205959

Homework Equations

$$ \log_{x} a \ means \ 10^x = a $$
$$ \log_{x} {a^n} = n \log_{x}{a} $$

The Attempt at a Solution

$$ \log_{10} {ax} \log_{10}{bx} + 1 = 0 $$
$$ \ means \log_{10} {ax} = 1 \log_{10}{bx} = -1 \ or \ vice \ versa $$
$$ \ for \log_{10} {ax} = 1 \ so \ ax \ = 10 \ and \ \log_{10}{bx} = -1 \ so \ bx \ = 0.1 \ or \ vice \ versa $$

No, not true. It just means that $\log_{10}(ax)\cdot \log_{10}(bx) = -1$. For example, one could be 1/2 and the other could be -2.
Work from this equation, and use the fact that $\log(mn) = \log(m) + \log(n)$.
I haven't worked the problem, but this is what I would do.

bx/ax = b/a = 0.1/10 = 0.01
or b/a = 10/0.1 = 100
so b/a > ?
how to make range for b/a ? how's the correlation with x?

 

[fresh_42]

What's the formula for $\log (p \cdot q)$? And what do you get, if you apply it to $\log(ax)\cdot \log(bx)$

 

[Ray Vickson]

Homework Statement

View attachment 205959

Homework Equations

$$ \log_{x} a \ means \ 10^x = a $$
$$ \log_{x} {a^n} = n \log_{x}{a} $$

The Attempt at a Solution

$$ \log_{10} {ax} \log_{10}{bx} + 1 = 0 $$
$$ \ means \log_{10} {ax} = 1 \log_{10}{bx} = -1 \ or \ vice \ versa $$
$$ \ for \log_{10} {ax} = 1 \ so \ ax \ = 10 \ and \ \log_{10}{bx} = -1 \ so \ bx \ = 0.1 \ or \ vice \ versa $$
bx/ax = b/a = 0.1/10 = 0.01
or b/a = 10/0.1 = 100
so b/a > ?
how to make range for b/a ? how's the correlation with x?

No, $\log_x a$ does not mean $10^x = a$; it means that $y = \log_x a$ is the solution of the equation $x^y = a$.

Anyway, your "solution" is all wrong; you are assuming things that need not be true.

Using the standard result that $\log_{10}(ax) = \log_{10} a + \log_{10} x$, etc., will help.

 

[Helly123]

ok, i get
(loga + log x) . (logb + log x) = -1
$$ \log_{10}{a} . \log_{10}{b} + 2\log_{10}{x} = -1$$

 

[Ray Vickson]

ok, i get
(loga + log x) . (logb + log x) = -1
$$ \log_{10}{a} . \log_{10}{b} + 2\log_{10}{x} = -1$$

Still wrong. Check your algebra.

 

[Helly123]

Oh

Still wrong. Check your algebra.

.. Loga.logb + (logx)^2 = -1

 

[fresh_42]

ok, i get
(loga + log x) . (logb + log x) = -1

Let's make it easier and forget about the logarithms for a moment. Say $a_l = \log a\, , \,x_l = \log x \; , \;b_l=\log b$.
Now what is $(a_l + x_l) \cdot (b_l +x_l) + 1 = 0\,$?

 

[Ray Vickson]

Oh

.. Loga.logb + (logx)^2 = -1

Are you working things out in detail, or are you just guessing? (Because you still have not gotten it right, although you are getting a bit closer.)

 

[Helly123]

Let's make it easier and forget about the logarithms for a moment. Say $a_l = \log a\, , \,x_l = \log x \; , \;b_l=\log b$.
Now what is $(a_l + x_l) \cdot (b_l +x_l) + 1 = 0\,$?

ab + x(a +b) + x^2 + 1 = 0
loga . log b + logx(log a + log b) + (logx)^2 + 1 = 0

 

[fresh_42]

ab + x(a +b) + x^2 + 1 = 0
loga . log b + logx(log a + log b) + (logx)^2 + 1 = 0

Yes, and now solve this equation and consider especially the sign of the two solutions and the expression under the root.

 

[Helly123]

I still don't get it. Can anyone give me hint?

 

[Dick]

I still don't get it. Can anyone give me hint?

It's a quadratic equation with $log(x)$ as the variable. Use the quadratic formula!

 

[Helly123]

let's say I want to solve the equation ab + x(a +b) + x^2 + 1 = 0
using abc formula : { -(a+b) +- root [ (a+b)^2 - 4(ab + 1) ] } / 2
{-a-b +- root [ a^2 + b^2 - 2ab -4 ] } /2
what am I suppose to do next?

 

[Helly123]

root1 * root2 = c/a
root 1 + root 2 = -b / a
so,
root 1 * root 2 = c/a = { ab + 1 } / 1 = loga + log b + log 10
what to do to find root 1 and 2 which is contains log a and log b and contains a and b?

 

[Ray Vickson]

let's say I want to solve the equation ab + x(a +b) + x^2 + 1 = 0
using abc formula : { -(a+b) +- root [ (a+b)^2 - 4(ab + 1) ] } / 2
{-a-b +- root [ a^2 + b^2 - 2ab -4 ] } /2
what am I suppose to do next?

You are supposed to figure out what the condition that at least one root is > 0 has to say about a and b.

 

[Dick]

let's say I want to solve the equation ab + x(a +b) + x^2 + 1 = 0
using abc formula : { -(a+b) +- root [ (a+b)^2 - 4(ab + 1) ] } / 2
{-a-b +- root [ a^2 + b^2 - 2ab -4 ] } /2
what am I suppose to do next?

If there is going to be a solution then the quantity in the square root needs to be nonnegative. Write down that condition and see if you can simplify it.