estrigaslju
2022-04-22
Answered

Solve in terms of b:

${\mathrm{log}}_{b}(1-3x)=3+{\mathrm{log}}_{b}x$

You can still ask an expert for help

Gianna Travis

Answered 2022-04-23
Author has **11** answers

Remember, any $\mathrm{log}$ to the base b, such as ${\mathrm{log}}_{b}\left(y\right)$ , is in its heart an exponent. Let us start, then, from your expression

${\mathrm{log}}_{b}\left(\frac{1-3x}{x}\right)=3.$

Raise b to the powers we see on each side. We get

${b}^{{\mathrm{log}}_{b}\left(\frac{1-3x}{x}\right)}={b}^{3}.$

The left-hand side simplifies greatly. We get

$\frac{1-3x}{x}={b}^{3}.$

The rest is elementary algebra. The above equation is (for$x\ne 0$ ) equivalent to

$1-3x={b}^{3}x,$

which is an easily solved linear equation.

Comment: There was no need to do the preliminary manipulation. We are told that

${\mathrm{log}}_{b}(1-3x)=3+{\mathrm{log}}_{b}x.$

Raise b to the power on the left-hand side, the right-hand side. We obtain

${b}^{{\mathrm{log}}_{b}(1-3x)}={b}^{3+{\mathrm{log}}_{b}x}.$

By the "laws of logarithms" this yields

$1-3x={b}^{3}x.$

Raise b to the powers we see on each side. We get

The left-hand side simplifies greatly. We get

The rest is elementary algebra. The above equation is (for

which is an easily solved linear equation.

Comment: There was no need to do the preliminary manipulation. We are told that

Raise b to the power on the left-hand side, the right-hand side. We obtain

By the "laws of logarithms" this yields

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