SOLUTION: Please help me solve this. 2log (27/8) base x =6 and 1/3log x base 10 =1

Question 1043577: Please help me solve this.
2log (27/8) base x =6
and
1/3log x base 10 =1
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
start with 1/3 * log(x) to the base of 10 = 1.

multiply both side of this equation by 3 to get log(x) to the base of 10 = 3.

this is true if and only if 10^3 = x

this makes x = 1000.

that's the solution to your second question.

now to your first.

2 * log(27/8) to the base of x = 6

divide both sides of this equation by 2 to get log(27/8) to the base of x = 3

this is true if and only if x^3 = 27/8

raise both sides of this equation to the 1/3 power to get x = (27/8)^(1/3)

solve for x to get x = 3/2.

you can confirm the solutions are correct by replacing x in the original equations with their respective values.

in the first problem, replace x with 3/2.

in the second problem, replace x with 1000.

to confirm the solution to the second equation, you can use the log base conversion formula of log(x) to the base of b is equal to log(x) to the base of 10 divided by log(b) to the base of 10.

the log function of your calculator is programmed to give you log(x) to the base of 10.

i usually represent log to the base of 10 as LOG.

to confirm the solution to your second problem, you would take 2 * log(27/8) to the base of 3/2 and calculate it using your calculator as 2 * LOG(27/8) / LOG(3/2).

your other option is to do the following:

start with 2 * log(27/8) to the base of 3/2 = 6.

divide both sides of the equation by 2 to get log(27/8) to the base of 3/2 = 3.

this is true if and only if (3/2)^3 = 27/8.

simplify to get 27/8 = 27/8.

since this is true, the solution is correct.

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