1) Find \(\log_{5 \sqrt{5}}(625) \)
2) If \(\log_{10}(x) = 3 + \log_{10}(y)\) then find \(\dfrac{x}{y} \)
3) Let a and b be positive real numbers so that \( \log_b(x^2) = 10\) Find \(\log_{\sqrt[3]{b}} \left( \frac{1}{x} \right) \)
4)Find \(\frac{a}{b}\) when \(2\log{(a -2b)} = \log(a) + \log(b)\)
+129849 (1) log 5 √5 (625) = log 5 √5 (5)^4
Using the change of base rule we can evaluate this in the following manner :
log 5^4 4 log 5 4 log 5 4
______ = ___________ = __________ = _____ = 4 (2/3) = 8 / 3
log 5 √5 log 5^(3/2) (3/2) log 5 (3/2)
(2) log x = 3 + log y subtract log y from both sides
log x - log y = 3
log (x / y) = 3
Note that we can write 3 as log 1000
So
log (x / y) = log 1000
So .... (x/y) = 1000
(3) log b x^2 = 10
2 logb x = 10 divide through by 2
log b x = 5
This says that b^5 = x
So we have that
log ∛b (1/ x) = log ∛b (1/b^5) = log b^(1/3) (b)^(-5) = (-5) log b^(1/3)(b) = -5 (3) = -15
(4) 2log(a - 2b) = log a + log b
log(a - 2b)^2 = log (ab)
This implies that
(a - 2b)^2 = ab simplify
a^2 - 4ab + 4b^2 = ab
a^2 - 5ab + 4b^2 = 0 factor as
(a - 4b) ( a - b) = 0
The first term gives us what we need.....setting it to 0, we get that
a - 4b = 0
a = 4b
a / b = 4
