Simplifying expressions with powers and logarithms

Text

Simplify using the properties of powers and logarithms (the result is a number):

• (a) $\log_a a^5 + \log_b\left(\dfrac{1}{b^3}\right) - \log_c \sqrt{c}$
• (b) $\log_{a^2} a^3 + \log_{1/a} a^4$
• (c) $\dfrac{\log_a (a^3 b^2) \cdot \log_b b^4}{\log_c c^3}$ (with $\log_a b=3$)
• (d) $\left(a^2\right)^{\log_a 2} - c^{\log_c (3) + 1}$
• (e) $\log_a (c^2 b^3)-\log_b(c/a)$ (with $\log_b a=2$ and $\log_c a=3$)

Solution

(a) $5+(-3)-\tfrac12=\tfrac32$.

(b) $\log_{a^2}a^3=\tfrac32$ and $\log_{1/a}a^4=-4$, so $\tfrac32-4=-\tfrac52$.

(c) $\log_a(a^3b^2)=3+6=9$; $\log_b b^4=4$; $\log_c c^3=3$: $\dfrac{9\cdot4}{3}=12$.

(d) $\left(a^2\right)^{\log_a 2}=2^2=4$; $c^{\log_c 3+1}=3c$. Result $4-3c$: it depends on $c$ (numeric only if the term is $c^{\log_c 3}=3$, giving $1$).

(e) $\log_a c=\tfrac13$, $\log_a b=\tfrac12$: $\log_a(c^2b^3)=\tfrac{13}{6}$; $\log_b(c/a)=-\tfrac43$. Total $\tfrac{21}{6}=\tfrac72$.

$\boxed{\text{(a) }\tfrac32;\ \text{(b) } -\tfrac52;\ \text{(c) } 12;\ \text{(d) } 4-3c;\ \text{(e) } \tfrac72}$