I got a logarithm question ?

Question

First question

given that log(base 3) 5 = 1.585 and log(base 2) 5 = 2.322. without using logarithm tables/calculator, find the value of log(base 5) 0.12

second question

5^x+2 =120 + 5^x

third question

log(base 5) x = log(base25) (3x+4)

can someone show me the work..

i got it wrong and it's bugging me to know how it's done..

Answer 1

Third Question"

look at the RHS:

let y = log(base 25) (3x + 4)

so 25^y = 3x + 4

but 25^y = (5^2)^y = 5^(2y) = 3x + 4

converting back to log form:

log(base 5) (3x + 4) = 2y = 2 log (base 25) (3x + 4)

(1/2) log (base 5) (3x + 4) = log (base 25) (3x + 4)

substitute this into the original

log (base 5) x = (1/2) log (base 5) (3x + 4)

2 log (base 5) x = log (base 5) (3x + 4)

log (base 5) (x^2) = log (base 5) (3x + 4)

now we have an equation with two logs with a common base; you can disregard the logs, and just work with the arguments:

x^2 = 3x + 4

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4 or x = -1

but x can't be -1 (domain of log x is x > 0)

so x = 4 is the only solution

check: log (base 5) 4 = log (base 25) (3*4 + 4) = log (base 25) 16?

log 4 / log 5 = log 16 / log 25?

note that log 16 = 2 log 4 (since 16 = 4^2) and log 25 = 2 log 5 (25 = 5^2)

so the RHS simplifies to log 4 / log 5, and this checks

Answer 2

First question:

The assumptions are wrong.

Log (base 3) 5 = 1.464973521 (quite different than what you typed)

Log (base 2) 5 = 2.321928095 (which agrees with what you typed)

I still can't figure out how to do it, but this may help someone else

Second question:

I'm assuming you meant:

5^(x+2) = 120 + 5^x

(5^x)*(5^2) = 120 + 5^x

25*(5^x) = 120 + 5^x

25*(5^x) - 5^x = 120

(5^x)*(25-1) = 120

5^x = 120/24

5^x = 5

5^x = 5^1

x = 1 (since the base "5" is the same, then the exponents must be equal)

Third question:

log (base 5) x = log (base 25) (3x+4)

Let's just work on the right hand side first:

log (base 25) (3x+4) = y

so 25^y = 3x + 4

or 5^(2y) = 3x + 4

which becomes:  log (base 5) (3x + 4) = 2y

or y = 1/2 log(base 5) (3x + 4)

finally, y = log (base 5) [(3x+4)^(1/2)]

substituting "y" back into the original question:

log (base 5) x = log (base 5) [(3x+4)^(1/2)]

Since the log and base are the same, then the argument inside the log functions must be the same:

x = [(3x+4)^(1/2)]

x^2 = 3x + 4

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4, -1

arguments inside the log function cannot result in negatives, so ignore the x = -1 solution.

The final answer is x = 4.

Answer 3

1)

log(base a) b = 1/[log(base b) a], so:

log(base 5) 3 = 1/1.585

log(base 5) 0.12 = log(base 5) 3 - 2*log(base 5) 5 = (1/1.585) - 2

2)

5^(x+2) = 120 + 5^x

25*(5^x) = 120 + 5^x

24*(5^x) = 120

5^x = 5

x = 1

3)

log(base 5) a = log(base 25) a^2, so:

log(base 5) x = log(base 25) (3x + 4)

log(base 25) x^2 = log(base 25) (3x + 4)

x^2 = 3x + 4

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4 (we throw out x = -1 because -1 doesn't make sense in a logarithm... unless you're working with complex numbers, but then log(x) isn't one-to-one so the above argument doesn't really work).

Answer 4

question 2

x = 1