Logarithm Equation: Rewriting log-x to log -x^2

[Heatherirving]

Hi

I wonder how I can rewrite log-x to log -x^2
View attachment 6584

 

[MarkFL]

Hello and welcome to MHB, Heatherirving! (Wave)

Let's consider that:

\(\displaystyle \log_x(a)=b\)

Can be translated from logarithmic to exponential form as:

\(\displaystyle a=x^b\)

Now, let's rewrite this as:

\(\displaystyle a=x^{2\frac{b}{2}}=\left(x^2\right)^{\frac{b}{2}}\)

Now, if we convert this to logarithmic form, we have:

\(\displaystyle \log_{x^2}(a)=\frac{b}{2}\)

And so we conclude:

\(\displaystyle b=\log_x(a)=2\log_{x^2}(a)\)

Or:

\(\displaystyle \log_x(a)=\log_{x^2}\left(a^2\right)\)

 

[Heatherirving]

Is it right to write the equation as:

(3-x)^2 = ( 8-3x-x^2)

 

[MarkFL]

Is it right to write the equation as:

(3-x)^2 = ( 8-3x-x^2)

Yes, but bear in mind what makes a meaningful logarithmic base. That is:

By definition of logarithmic functions, we know that the base of a logarithmic function is a positive number excluding $x=1$. ;)

 

[Heatherirving]

Through my calculation, I got that X (1) = 1 and X (2) = 0,5 According to the question, you should also indicate the largest (real) solution.

is not 1 a larger real solution than 0,5

 

[MarkFL]

Through my calculation, I got that X (1) = 1 and X (2) = 0,5 According to the question, you should also indicate the largest (real) solution.

is not 1 a larger real solution than 0,5

We do get \(\displaystyle x\in\left\{\frac{1}{2},1\right\}\), from the resulting equation, but we have to discard $x=1$ since it makes no sense as a logarithmic base, as it is used in the original equation. So, the only valid solution, and therefore the largest is:

\(\displaystyle x=\frac{1}{2}\)

 

[Heatherirving]

thanks for the help :)