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logx^3logx^2log14 is 3logx2logxlog14 this equals 6 log14 (logx)^2 So for example, if y=6log14(logx)^2 the log x = square root of (y/6(log14))

Q: Logx 3 logx 2 log 14?

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Without tables or calculators, this kinda question is hard, so if there is a non-calculator question involving logs, just chuck numbers in. I find it helpful to rearrange the log into a power problem e.g. log2(x) = 8 --------> x = 28 = 256 or log2(8) = x ----------> 2x = 8 = 3 (by observation) or logx(8) = 2 ----------> x2 = 8, x = (+/-)(root 8) = (+/-)(2root2)

a log is the 'undo-er' of powers, kind of like division is the 'undo-er' of multiplication. EX: 102 = 100, then log10(100) = 2 103 = 1000, then log10(1000) = 3, in this example, we are using log base 10, this is a default base and sometimes isn't even wirten. e is probably the most common base but log base e is more simply called the natural log, or ln. so in general: logx(m) = N means that xN = m so log5(125) = 3 because 53 = 125.

You will need 14 two's multiplied together to equal 16384. the answer to this can be found by log2(16384) = 14. Since most calculators don't have log base 2, you can do this: log(16384)/log(2) = 14. You can use the 'base 10' log or natural log [ln(16384)/ln(2) = 14]

0.03

log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1

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log(100x) can be written as log100 + logx. This =2+logx

logx^2=2 2logx=2 logx=1 10^1=x x=10

logx = 2 so x = 10logx = 102 = 100 ie x = 100.

Without brackets, there are many ambiguities in the question. 4 = 2/3*log(x) - 0.9 4.9*3/2 = logx 7.35 = logx so x = 107.35 However, the original equation could also have referred to 4 = 2/3*log(x-0.9) or 4 = 2/[3*log(x)] - 0.9 or 4 = 2/[3*log(x-0.9)]

11

(1/4)x^2(2logx-1) + c

We must assume that the question is asking us to determine the value of 'x'.log(7) + log(x) = 2log(7x) = 27x = 102 = 100x = 100/7 = 14.2857 (rounded)

2logx-log5=-2 logx^2-log5=-2 log(x^2 / 5)=-2 x^2 / 5 = 10^-2 x^2=5(1/100) x^2=1/20 x=√(1/20)

We can't see the parentheses, and there are at least two ways to read this.Here are solutions for the most likely two:--------------------------------------------------------log(x) - 3 + log(x) - 2 = log(2x) + 24Add 5 to each side:log(x) + log(x) = log(2x) + 29Subtract log(2x) from each side:log(x) + log(x) - log(2x) = 29Combine the logs on the left side, and massage:log( x2/2x ) = log( x/2 ) = 29Take the antilog of each side:x/2 = 1029Multiply each side by 2:x = 2 x 1029------------------------------------------------log(x - 3) + log(x - 2) = log(2x + 24)Combine logs on the left side:log[ (x-3) (x-2) ] = log(2x + 24)Take antilog of each side:(x-3) (x-2) = 2x + 24Expand the left side:x2 - 5x +6 = 2x + 24Subtract (2x+24) from each side:x2 - 7x - 18 = 0Factor:(x - 9) (x + 2) = 0Whence:x = 9x = -2We have to discard the solution [ x = -2 ] because one term in the equationis log(x-2).If 'x' were -2 then we'd have log(-4) but negative numbers don't have logs.

128 = x7 (we see that x > 0, since x is raised to an odd power) 27 = x7 (this is true only when x = 2) 2 = x Remember that a logarithm is an exponent. The statement 128 = x7 is equivalent to logx 128 = 7. logx 128 = 7 log 128/log x = 7 (or reverse the both sides) log x/log 128 = 1/7 (multiply by log 128 to both sides) log x = log128/7 (or use base 10 for the logarithm) log10 x = log128/7 (write the equivalent statement) 10log 128/7 = x 2 = x or use the natural log, ln. 128 = x7 ln 128 = ln x7 ln 128 = 7ln x (ln 128)/7 = ln x e(ln128)/7 = elnx (elnx = x ) 2 = x

assuming the bases are the same, you can factor the log out, so you get:x^2 = 2 + 3x-4 (note that 2logx can be rewritten as logx^2 )x^2 - 3x +2 = 0You can factor this out: (x-1)(x-2) = 0so x must be one or x is two --> X=1 V X=2

Without tables or calculators, this kinda question is hard, so if there is a non-calculator question involving logs, just chuck numbers in. I find it helpful to rearrange the log into a power problem e.g. log2(x) = 8 --------> x = 28 = 256 or log2(8) = x ----------> 2x = 8 = 3 (by observation) or logx(8) = 2 ----------> x2 = 8, x = (+/-)(root 8) = (+/-)(2root2)