Question and Answers Forum

LogarithmsQuestion and Answers: Page 4

Question Number 182904 Answers: 1 Comments: 0

Question Number 182737 Answers: 1 Comments: 0

Question Number 182632 Answers: 1 Comments: 0

If log _(11) (3p)=log _(13) (q+6p) = log _(143) (q^2 ) find (p/q).

$$\:\:\mathrm{If}\:\mathrm{log}\:_{\mathrm{11}} \left(\mathrm{3p}\right)=\mathrm{log}\:_{\mathrm{13}} \left(\mathrm{q}+\mathrm{6p}\right)\:=\:\mathrm{log}\:_{\mathrm{143}} \left(\mathrm{q}^{\mathrm{2}} \right) \\ $$$$\:\mathrm{find}\:\frac{\mathrm{p}}{\mathrm{q}}. \\ $$

Question Number 181154 Answers: 1 Comments: 0

(log_(15) 5)^2 +(log_(15) 3)(log_(15) 75)=?

$$\left({log}_{\mathrm{15}} \mathrm{5}\right)^{\mathrm{2}} +\left({log}_{\mathrm{15}} \mathrm{3}\right)\left({log}_{\mathrm{15}} \mathrm{75}\right)=? \\ $$$$ \\ $$

Question Number 180014 Answers: 2 Comments: 0

Given 80^a = 5 and 80^b = 2 then 25^((1−a−2b)/(1+a−4b)) =?

$$\:\:\mathrm{Given}\:\mathrm{80}^{{a}} \:=\:\mathrm{5}\:\mathrm{and}\:\mathrm{80}^{{b}} \:=\:\mathrm{2} \\ $$$$\:\mathrm{then}\:\mathrm{25}^{\frac{\mathrm{1}−{a}−\mathrm{2}{b}}{\mathrm{1}+{a}−\mathrm{4}{b}}} \:=?\: \\ $$

Question Number 179722 Answers: 1 Comments: 2

1000^(500 x) = (√(200))

$$\mathrm{1000}^{\mathrm{500}\:{x}} \:=\:\sqrt{\mathrm{200}} \\ $$

Question Number 179679 Answers: 1 Comments: 0

∫_2 ^4 tan (x)

$$\int_{\mathrm{2}} ^{\mathrm{4}} \mathrm{tan}\:\left({x}\right) \\ $$

Question Number 177361 Answers: 1 Comments: 0

4(16^((x+4)) × 5.2^(2x) =13

$$\mathrm{4}\left(\mathrm{16}^{\left({x}+\mathrm{4}\right)} \:×\:\mathrm{5}.\mathrm{2}^{\mathrm{2}{x}} =\mathrm{13}\right. \\ $$

Question Number 177360 Answers: 1 Comments: 0

1/2 log_4 36 ×log_6 64

$$\mathrm{1}/\mathrm{2}\:{log}_{\mathrm{4}} \mathrm{36}\:×{log}_{\mathrm{6}} \mathrm{64} \\ $$

Question Number 177113 Answers: 0 Comments: 9

X and Y are playing a game. Initially there are three bundles of matches, consisting of 2021, 2022 and 2023 pieces. Each player in his turn chooses a bundle B and removes a positive number of the matches of B such that the number of pieces of bundles still form an arithmetic sequence. The player who cannot do a legal move loses. Determine which player has the winning strategy.

$${X}\:{and}\:{Y}\:{are}\:{playing}\:{a}\:{game}.\: \\ $$$${Initially}\:{there}\:{are}\:{three}\:{bundles}\:{of}\: \\ $$$${matches},\:{consisting}\:{of}\:\mathrm{2021},\:\mathrm{2022}\: \\ $$$${and}\:\mathrm{2023}\:{pieces}.\:{Each}\:{player}\:{in}\:{his}\: \\ $$$${turn}\:{chooses}\:{a}\:{bundle}\:{B}\:{and}\:{removes}\: \\ $$$${a}\:{positive}\:{number}\:{of}\:{the}\:{matches}\:{of}\:{B}\: \\ $$$${such}\:{that}\:{the}\:{number}\:{of}\:{pieces}\:{of}\: \\ $$$${bundles}\:{still}\:{form}\:{an}\:{arithmetic}\: \\ $$$${sequence}.\:{The}\:{player}\:{who}\:{cannot}\:{do}\:{a}\: \\ $$$${legal}\:{move}\:{loses}.\:{Determine}\:{which}\: \\ $$$${player}\:{has}\:{the}\:{winning}\:{strategy}. \\ $$

Question Number 176332 Answers: 1 Comments: 1

f((x/(f(y))))= (x/y) solve for x and y

$${f}\left(\frac{{x}}{{f}\left({y}\right)}\right)=\:\frac{{x}}{{y}} \\ $$$${solve}\:{for}\:{x}\:{and}\:{y} \\ $$

Question Number 176310 Answers: 1 Comments: 0

a^(lna) =b^(lnb) a−b=1 solve for a and b

$${a}^{{lna}} ={b}^{{lnb}} \\ $$$${a}−{b}=\mathrm{1} \\ $$$${solve}\:{for}\:{a}\:{and}\:{b} \\ $$

Question Number 175104 Answers: 1 Comments: 0

log _3 (a+1)=log _4 (a+8) a=?

$$\:\:\mathrm{log}\:_{\mathrm{3}} \left({a}+\mathrm{1}\right)=\mathrm{log}\:_{\mathrm{4}} \left({a}+\mathrm{8}\right) \\ $$$$\:{a}=? \\ $$

Question Number 175040 Answers: 1 Comments: 0

hello, please, someone help me to correct the equation? It's typed wrong and I can't find where Solve: 5,76[((log_a ((√(log _b ((√a))))))/(log((√(log(a)))))) + log_(log (a)) (2)]((log _2 (x)))^(1/5) + ((log_2 (x))/(25)) = [log _2 (x)]^(3/5) Answers x_1 =1 , x_2 =2^(243) , x_3 =2^(−243) , x_4 =2^(1024) , x_5 =2^(−1024)

$$ \\ $$hello, please, someone help me to correct the equation? It's typed wrong and I can't find where $${Solve}: \\ $$$$\mathrm{5},\mathrm{76}\left[\frac{\mathrm{log}_{{a}} \left(\sqrt{\mathrm{log}\:_{{b}} \left(\sqrt{{a}}\right)}\right)}{\mathrm{log}\left(\sqrt{\mathrm{log}\left({a}\right)}\right)}\:+\:\mathrm{log}_{\mathrm{log}\:\left({a}\right)} \left(\mathrm{2}\right)\right]\sqrt[{\mathrm{5}}]{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}\:+\:\frac{\mathrm{log}_{\mathrm{2}} \left({x}\right)}{\mathrm{25}}\:=\:\left[\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\right]^{\frac{\mathrm{3}}{\mathrm{5}}} \\ $$$${Answers} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}\:,\:{x}_{\mathrm{2}} =\mathrm{2}^{\mathrm{243}} \:,\:{x}_{\mathrm{3}} =\mathrm{2}^{−\mathrm{243}} \:,\:{x}_{\mathrm{4}} =\mathrm{2}^{\mathrm{1024}} \:,\:{x}_{\mathrm{5}} =\mathrm{2}^{−\mathrm{1024}} \\ $$

Question Number 174940 Answers: 0 Comments: 1

f(x)=(log3^x −2log3).(x^2 −1) let

$${f}\left({x}\right)=\left({log}\mathrm{3}^{{x}} −\mathrm{2}{log}\mathrm{3}\right).\left({x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${let} \\ $$

Question Number 174287 Answers: 2 Comments: 0