ln^2 (4) is not the same as (ln4)^2..... Take note of this...

Some calculators and computing engines do accept ln² (4) and process it as (ln 4)².

But for clarity, always use (ln 4)² in writing.

Avoid writing it like trigo eg. cos² x , cosec² π

https://socratic.org/questions/is-lnx-2-equivalent-to-ln-2-x

ln² x is actually an old way of writing (ln x)² so it depends on whether your marker/examiner accepts such notation.

Zwen, from your third step and following the notation you use: (I presume your lecturer uses this)

(Also remember that (ln 4)(ln x) ≠ ln 4x
ln 4x = ln 4 + ln x)


Working :


(ln 4)(ln 4 + ln x) = (ln 5)(ln 5 + ln x)

ln² 4 + (ln 4)(ln x) = ln² 5 + (ln 5)(ln x)

ln² 4 - ln² 5 = (ln 5)(ln x) - (ln 4)(ln x)

(ln 4 - ln 5)(ln 4 + ln 5) = (ln 5 - ln 4)(ln x)


[Recall the property a² - b² = (a + b)(a - b) ]

-(ln 5 - ln 4)(ln 4 + ln 5) = (ln 5 - ln4)(ln x)

-(ln 5 - ln 4)(ln 4 + ln 5) / (ln 5 - ln4) = ln x

-(ln 4 + ln 5) = ln x

-(ln (4 × 5)) = ln x

-ln 20 = ln x

ln 20-¹ = ln x

ln 1/20 = ln x

Since both log bases are the same on both sides, the arguments are equal

x = 1/20 (or 0.05)


No calculator is needed , which Fergus used.

Thank you all! Appreciate it.