Notes: `10^(3x-10)=(1/100)^(6x-1)` Solve the equation.

Saturday, 11 January 2014

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left(\frac{{1}}{{100}}\right)}}^{{{6}{x}-{1}}}$ Solve the equation.

To evaluate the given equation $\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left(\frac{{1}}{{100}}\right)}}^{{{6}{x}-{1}}}$ , we may apply $\displaystyle{100}={{10}}^{{2}}$ . The equation becomes:

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left(\frac{{1}}{{{10}}^{{2}}}\right)}}^{{{6}{x}-{1}}}$

Apply Law of Exponents: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{n}}}$ .

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left({{10}}^{{-{2}}}\right)}}^{{{6}{x}-{1}}}$

Note: $\displaystyle\frac{{1}}{{100}}={{10}}^{{-{2}}}$

Apply Law of Exponents: $\displaystyle{{\left({{x}}^{{n}}\right)}}^{{m}}={{x}}^{{{n}\cdot{m}}}$ .

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{10}}^{{{\left(-{2}\right)}\cdot{\left({6}{x}-{1}\right)}}}$

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{10}}^{{-{12}{x}+{2}}}$

Apply the theorem: If $\displaystyle{{b}}^{{x}}={{b}}^{{y}}$ then $\displaystyle{x}={y}$ , we get:

$\displaystyle{3}{x}-{10}=-{12}{x}+{2}$

Add $\displaystyle{12}{x}$ on both sides of the equation.

$\displaystyle{3}{x}-{10}+{12}{x}=-{12}{x}+{2}+{12}{x}$

$\displaystyle{15}{x}-{10}={2}$

Add $\displaystyle{10}$ on both sides of the equation.

$\displaystyle{15}{x}-{10}+{10}={2}+{10}$

$\displaystyle{15}{x}={12}$

Divide both sides by $\displaystyle{15}$ .

...

To evaluate the given equation $\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left(\frac{{1}}{{100}}\right)}}^{{{6}{x}-{1}}}$ , we may apply $\displaystyle{100}={{10}}^{{2}}$ . The equation becomes:

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left(\frac{{1}}{{{10}}^{{2}}}\right)}}^{{{6}{x}-{1}}}$

Apply Law of Exponents: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{n}}}$ .

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left({{10}}^{{-{2}}}\right)}}^{{{6}{x}-{1}}}$

Note: $\displaystyle\frac{{1}}{{100}}={{10}}^{{-{2}}}$

Apply Law of Exponents: $\displaystyle{{\left({{x}}^{{n}}\right)}}^{{m}}={{x}}^{{{n}\cdot{m}}}$ .

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{10}}^{{{\left(-{2}\right)}\cdot{\left({6}{x}-{1}\right)}}}$

$\displaystyle{{10}}^{{{3}{x}-{10}}}={{10}}^{{-{12}{x}+{2}}}$

Apply the theorem: If $\displaystyle{{b}}^{{x}}={{b}}^{{y}}$ then $\displaystyle{x}={y}$ , we get:

$\displaystyle{3}{x}-{10}=-{12}{x}+{2}$

Add $\displaystyle{12}{x}$ on both sides of the equation.

$\displaystyle{3}{x}-{10}+{12}{x}=-{12}{x}+{2}+{12}{x}$

$\displaystyle{15}{x}-{10}={2}$

Add $\displaystyle{10}$ on both sides of the equation.

$\displaystyle{15}{x}-{10}+{10}={2}+{10}$

$\displaystyle{15}{x}={12}$

Divide both sides by $\displaystyle{15}$ .

$\displaystyle\frac{{{15}{x}}}{{15}}=\frac{{12}}{{15}}$

$\displaystyle{x}=\frac{{12}}{{15}}$

Simplify.

$\displaystyle{x}=\frac{{4}}{{5}}$

Checking: Plug-in $\displaystyle{x}=\frac{{4}}{{5}}$ on $\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left(\frac{{1}}{{100}}\right)}}^{{{6}{x}-{1}}}$ .

$\displaystyle{{10}}^{{{3}\cdot{\left(\frac{{4}}{{5}}\right)}-{10}}}=?{{\left(\frac{{1}}{{100}}\right)}}^{{{6}\cdot{\left(\frac{{4}}{{5}}\right)}-{1}}}$

$\displaystyle{{10}}^{{\frac{{12}}{{5}}-{10}}}=?{{\left(\frac{{1}}{{100}}\right)}}^{{\frac{{24}}{{5}}-{1}}}$

$\displaystyle{{10}}^{{\frac{{12}}{{5}}-\frac{{50}}{{5}}}}=?{{\left(\frac{{1}}{{100}}\right)}}^{{\frac{{24}}{{5}}-\frac{{5}}{{5}}}}$

$\displaystyle{{10}}^{{\frac{{-{38}}}{{5}}}}=?{{\left(\frac{{1}}{{100}}\right)}}^{{\frac{{19}}{{5}}}}$

$\displaystyle{{10}}^{{\frac{{-{38}}}{{5}}}}=?{{\left({{10}}^{{-{2}}}\right)}}^{{\frac{{19}}{{5}}}}$

$\displaystyle{{10}}^{{\frac{{-{38}}}{{5}}}}=?{{10}}^{{{\left(-{2}\right)}\cdot\frac{{19}}{{5}}}}$

$\displaystyle{{10}}^{{\frac{{-{38}}}{{5}}}}={{10}}^{{\frac{{-{38}}}{{5}}}}$ TRUE

Thus, the $\displaystyle{x}=\frac{{4}}{{5}}$ is the real exact solution of the equation $\displaystyle{{10}}^{{{3}{x}-{10}}}={{\left(\frac{{1}}{{100}}\right)}}^{{{6}{x}-{1}}}$ .