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What is the solution of log 2x – 3^125 = 3 ?
x = 1 over 3
x = 1
x = 7 over 3
x = 4
 one year ago
 one year ago
What is the solution of log 2x – 3^125 = 3 ? x = 1 over 3 x = 1 x = 7 over 3 x = 4
 one year ago
 one year ago

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telltoamitBest ResponseYou've already chosen the best response.0
log basex a = b implies x^a = b
 one year ago

WiredBest ResponseYou've already chosen the best response.0
Is this correct? \[\Large \log (2x3^{125}) = 3\]
 one year ago

telltoamitBest ResponseYou've already chosen the best response.0
nah the base is 2x  3
 one year ago

WiredBest ResponseYou've already chosen the best response.0
@telltoamit Only @gioanny39x can say that for sure. Gio, you need to confirm if I have the equation correct or not. Also, your formula is wrong. \[\Large log_B(X) = Y\] is the same as \[\Large X = B^{Y}\] Example: \[\Large log_10(1000) = 3\] B = 10 X = 1000 Y = 3 Using the 2nd form of the equation: \[\Large X = B^{Y}\] \[\Large 1000 = 10^{3}\]
 one year ago

ash2326Best ResponseYou've already chosen the best response.0
\[\log_{2x3} 125=3\] Is this your question @gioanny39x ?
 one year ago

telltoamitBest ResponseYou've already chosen the best response.0
only this gives the answer in options lol
 one year ago

gioanny39xBest ResponseYou've already chosen the best response.0
dw:1343929733059:dw
 one year ago

WiredBest ResponseYou've already chosen the best response.0
Ok, so: B = 2D3 (easier to use D to not confuse with X) X = 125 Y = 3 \[\Large X = B^{Y}\] \[\Large 125 = (2D3)^{3}\] Solve for D. Hint: 125 is the cube of a single digit number.
 one year ago

gioanny39xBest ResponseYou've already chosen the best response.0
(2d3)(2d3)(2d3) 4d^26d6d+9 4d^212d+9(2d3) 8d^312d^2 24d^2+36d 28d27 125=8d^3=36d^2+64d+27?
 one year ago

WiredBest ResponseYou've already chosen the best response.0
You're creating too much work for yourself. What is the cube root of 5? Hint: \[\Large X^{3} = (15Y+5)^{3}\] \[\Large X = (15Y+5)\] If you have two items that when cubed equal each other, the items equal each other.
 one year ago
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