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Log questions.

a.) What is x, for which;

3^(x-2)=2^x


b.) Solve;

lg(5x+8) + lg(2x+1) - 2lg(x+1) = 1

:confused:

Many thanks.
Reply 1
Avatar for Mikey™
Mikey™
Slice'N'Dice
a.) What is x, for which;

3^(x-2)=2^x
log 3^(x-2) = log 2^x
(x-2)log 3 = x log 2

b.) Solve;

lg(5x+8) + lg(2x+1) - 2lg(x+1) = 1

:confused:

Many thanks.


3^(x-2)=2^x
log 3^(x-2) = log 2^x
(x-2)log 3 = x log 2
(x-2)/x = (log 2)/(log 3)
1 - 2/x = 0.63
2/x = 0.369
x = 5.42 (3s.f.)
Reply 2
Avatar for Slice'N'Dice
Slice'N'Dice
OP
2
Cheers man, appriciate it. "a.)" was so basic!
Reply 3
Avatar for Fermat
Fermat
8
Slice'N'Dice

b.) Solve;

lg(5x+8) + lg(2x+1) - 2lg(x+1) = 1

:confused:

Many thanks.

Taking lg to mean log to the base 2, then lg(2) = 1

lg(5x+8) + lg(2x+1) - 2lg(x+1) = lg(2)
(5x+8)(2x+1)/(x+1)² = 2 (can you see how I get this ?)

multiplying out the brackets and cross-multiplyinbg,

10x² + 21x + 8 = 2x² + 4x + 2
8x² + 17x + 6 = 0
x = -17 ± √((17)² - 4.8.6) / 2.8
x = -17 ± √(97) / 16
x = -0.447
========
There is another solution, x = -1.678, but the domain of x in the log terms is such that x > -8/5 = -1.6, meaning that x = -1.678 is too low and so is ignored.

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