Janiah Parks

2022-10-10

Logarithmic Equation: Solve for x

$${\mathrm{log}}_{3x}81=2$$

How would I go about solving this? This is what I tried:

$${\mathrm{log}}_{3x}81=2$$

$$\frac{\mathrm{log}81}{\mathrm{log}3+\mathrm{log}x}=2$$

Where do I go from here?

If I isolate $\mathrm{log}x$ on one side, how do I get rid of the log?

$${\mathrm{log}}_{3x}81=2$$

How would I go about solving this? This is what I tried:

$${\mathrm{log}}_{3x}81=2$$

$$\frac{\mathrm{log}81}{\mathrm{log}3+\mathrm{log}x}=2$$

Where do I go from here?

If I isolate $\mathrm{log}x$ on one side, how do I get rid of the log?

Jaylyn George

Beginner2022-10-11Added 6 answers

${\mathrm{log}}_{3x}(81)=2$ is equivalent to

$$(3x{)}^{2}=9{x}^{2}=81$$

by the definition of the logarithm.

$$9{x}^{2}=81\iff {x}^{2}=9$$

This gives solutions $x=3$ and $x=-3$, but only $x=3$ is a solution, since the base of a logarithm must be greater than zero.

$$(3x{)}^{2}=9{x}^{2}=81$$

by the definition of the logarithm.

$$9{x}^{2}=81\iff {x}^{2}=9$$

This gives solutions $x=3$ and $x=-3$, but only $x=3$ is a solution, since the base of a logarithm must be greater than zero.

emmostatwf

Beginner2022-10-12Added 2 answers

For any two real numbers b and x where b is positive and $b\ne 1$,

$$y={b}^{z}\iff z={\mathrm{log}}_{b}(y)$$

so for ${\mathrm{log}}_{3x}81=2$ we have

$$(3x{)}^{2}=81=(3\cdot 3{)}^{2}\Rightarrow x=3$$

$$y={b}^{z}\iff z={\mathrm{log}}_{b}(y)$$

so for ${\mathrm{log}}_{3x}81=2$ we have

$$(3x{)}^{2}=81=(3\cdot 3{)}^{2}\Rightarrow x=3$$

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