Solving Equations Using Logarithms

WMA12/01/IAL/Edexcel/P2/Oct 2019/Q7 (Logarithms, Series)

Given​​ loga⁡b = k​​ find, in simplest form in terms of k,

• loga⁡ab

(2)

• loga⁡a2bloga⁡b3

(2)

• ∑n=150k+loga⁡bn 

(3)

SOLUTION​​ 

i- Usng the laws of log to solve the expression.​​ 

Applying the division rule.

loga⁡ab=loga⁡a-loga⁡b

Substituting ​​ loga⁡b=k,

loga⁡ab=loga⁡a12-k

Applying the power rule now.

loga⁡ab=12loga⁡a-k

Remember,​​ loga⁡a=1

loga⁡ab=12(1)-k 

ii- Applying the multiplication rule first.

loga⁡a2bloga⁡b3=loga⁡a2+loga⁡b3loga⁡b

Applying power rule now, to bring 2 down as the coefficient and Substituting ​​ loga⁡b=k.​​ 

=2loga⁡a+k3k

Remember,​​ loga⁡a=1

=2(1)+k3k

loga⁡a2bloga⁡b3=2+k3k

iii- Applying the power rule to bring n down as the coefficient.​​ 

∑n=150k+loga bn=∑n=150(k+nloga⁡b)

Since​​ loga⁡b=k, the expression becomes​​ 

=∑n=150k+nk

=∑n=150k1+n

On expanding it becomes

∑n=150k1+n=2k+3k+4k+… +51k

This gives us the arithmetic series, with​​ a=2k, and​​ l=51k,​​ 

Sn=n2a+l

S50=502 2k+51k=25 x 53k

S50=1325k