Question

Help needed for logarithmic equations , thanks in advance :)

Answer 1

\[\log \sqrt{x} = \log1 - 2\log3\]

Answer 2

$$
\log \sqrt{x} = \log1 - 2\log3\\
\log1 - 2\log3=0-2\log3\\
$$
Now raise the terms to the power 10:

$$
10^{\log \sqrt{x}}=\sqrt{x}=10^{-2\log3}\\
$$
Now square both sides
$$
x=\left (10^{-2\log3}\right )^2=10^{-4\log3}
$$
That's it!

Answer 3

perfect ! thank you very much @ybarrap :)

Answer 4

When you subtract logs the equation becomes this:
\[\log_{\sqrt{x}} = \frac{ \log 1 }{ 2\log3 }\]

Answer 5

\[\log \sqrt{x} = \frac{ \log1 }{ 2 \log 3 }\]

Answer 6

log 1 = ?

Answer 7

Here's a better way!

Solve for x over the real numbers:
\(\log(\sqrt{x}) = -2 \log(3)\)

Simplify logarithmic terms on the right hand side.
\(-2 \log(3) = \log(1/3^2) = \log(1/9):\)
\(\log(\sqrt{x}) = \log(1/9)\)

Eliminate the logarithm from the left hand side.
Cancel logarithms by taking exp of both sides:
\(\sqrt{x} = 1/9\)

Eliminate the square root on the left hand side.
Raise both sides to the power of two:
Answer: \( x = 1/81\)

Answer 8

Perfect alternative @ybarrap thank you :)

Thank you for your help as well @nelsonjedi :)