Solving a simple logarithmic equation?

Question

Solve the equation log (base 27) x = 1 - log (base27) (x-.0.4)

Please explain the solution to me. Choosing best answer! :)

Thanks in advance!

Answer 1

First you want to put the logarithms together to get rid of the log all together

log (base 27) x + log (base27) (x-.0.4) = 1

Adding two separate logs together, you just multiply the two quantities together:

log (base 27) [x (x - 0.4)] = 1

Now get rid of the log by using 27 as an exponent:

27^{ log (base 27) [x (x - 0.4)] } = 27^1

x (x - 0.4) = 27

Now solve for x from here:

x² - 0.4x - 27 = 0

5x² - x - 135 = 0

(5x - 27)(x + 5) = 0

5x - 27 = 0

5x = 27

x = 27/5

x + 5 = 0

x = -5

Plug in the two quantities, 27/5 & -5, and you'll find that -5 does not work because logarithms cannot compute negative numbers.

Therefore the answer is x = 27/5.

Answer 2

log 27 x + log 27 (x- 0.4) = 1

When adding logs, multiply

log 27 (x^2- .4x) = 1

(x^2 - .4x) = 27

5x^2 - 2x = 135

5x^2 - 2x - 135 = 0

(5x-27)(x+5)

x= -5 or (27/5)

x cannot be -5 in a log, so...

x= 27/5

Answer 3

log27 ( x ) = 1 - log27 ( x - 0.4 )

log27 ( x ) + log27 ( x - 0.4) = 1

Use the product property of logarithms.

log27 ( (x) (x - 0.4) ) = 1

log27 ( x² - 0.4x ) = 1

Use the inverse property of logarithms.

27^( log27 ( x² - 0.4x ) ) = 27^1

x² - 0.4x = 27

x² - 0.4x - 27 = 0

Using the quadratic equation I get

x = 5.4, x = -5

Can't take -5 of log, therefore

x = 5.4

Answer 4

log_27 x = 1 - log_27 (x - 0.4)

log_27 x + log_27 (x - 0.4) = 1

log_27 x(x - 0.4) = 1 .................multiplication rule

x(x - 0.4) = 27 .........................."undo" the log

x² - 0.4x - 27 = 0

from here, if your mistyped 0.4 is right, just use quadratic formula.