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Answer 1 · 2017-04-05T08:40:34

Solution to a) $\to x = \frac{1}{2}$ $->x=1/2$

Solution to a)

Adding logs reflects the action of multiplication of the source values.

So $ln (x - 1) + ln (3) = ln (x)$ $ln(x-1)+ln(3)=ln(x)$ is the same as: $ln (3 [x - 1]) = ln (x)$ $ln( 3[x-1])=ln(x)$

Using an example: note that ${ln}^{- 1} (a) = a$ $ln^(-1)(a)=a$

So ${ln}^{- 1} (3 [x - 1]) = {ln}^{- 1} (x)$ $" "ln^(-1)(3[x-1])" "=" "ln^(-1)(x)$

$3 (x - 1) = x$ $" "3(x-1)" "=" "x$

$3 x - 1 = x$ $" "3x-1=x$

$2 x = 1$ $" "2x=1$

$x = \frac{1}{2}$ $" "x=1/2$

Answer 2 · 2017-04-05T08:54:12

Solution to b) $" "x~~8.773...$ to 3 decimal places

Solution to b)

Using an example: note that $" "2ln(a) -> ln(a^2)$

So we have:

$ln[(5x)^3]-ln(x^2)=7$

$ln[(125x^3)/x^2]=7$

$ln(125x)=7$

$ln^(-1)(125x)=ln^(-1)(7) larr" not convinced this will work!"$

Trying something else!

$ln(125)+ln(x)=7$

$ln(x)=7-ln(125)$

$x=ln^(-1)(x)= ln^(-1)[color(white)(2/2)7-ln(125)color(white)(2/2)]$

$x~~8.773...$ to 3 decimal places