## Logarithm Help!! solve e^x-4 = 2, 4lnx = -2, n(2x-1)^2 = 4,

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### Logarithm Help!! solve e^x-4 = 2, 4lnx = -2, n(2x-1)^2 = 4,

Hi all,

I don't have time to introduce myself so I will get right to my question.

I would highly appreciate it if anyone could solve these out for me with some explanation, I have a test tomorrow and my Holt TX book is worthless.

(e is not a variable, it is the function and ^ denotes "to the power of", ln = natural log, b=# denotes the subscript)

1. e^x-4 = 2

2. 5e^6x+3 = 0.1

3. 4lnx = -2

4. ln(2x-1)^2 = 4

5. logbase=2(12)

6. The amount of money in a bank account can be expressed by the exponential equation A=300(1.005)^12(t) where A is the amount in dollars and t is the time in years. About how many years will it take for the amount in the account to be more than $900. RudyTheRocka Posts: 1 Joined: Thu Feb 26, 2009 4:01 am Sponsor RudyTheRocka wrote:I would highly appreciate it if anyone could solve these out for me with some explanation... For loads of on-topic worked examples, please review these lessons on solving exponential equations and solving log equations. Once you've learned the basic terms and techniques.... RudyTheRocka wrote:1. e^x-4 = 2 As posted, the equation is: . . . . .$\mbox{a. }\, e^x\, -\, 4\, =\, 2$ Is this correct, or is the equation as follows? . . . . .$\mbox{b. }\, e^{x\, -\, 4}\, =\, 2$ RudyTheRocka wrote:2. 5e^6x+3 = 0.1 Same question: Which of the following, if either, is correct? . . . . .$\mbox{a. }\, 5 e^{6x}\, +\, 3\, =\, 0.1$ . . . . .$\mbox{b. }\, 5 e^{6x\, +\, 3}\, =\, 0.1$ RudyTheRocka wrote:3. 4lnx = -2 Divide through by the 4 to get the log by itself. Then convert the log equation into its equivalent exponential form. This will give you "x=". How, and how much, you simplify will vary with the textbook and instructor, so check your class notes for advice in this area. For instance, some instructors really "care" about getting rid of negative exponents, while others are fine with them. RudyTheRocka wrote:4. ln(2x-1)^2 = 4 Is the log squared, or is the argument of the log squared? In other words, which of the following is correct? . . . . .$\mbox{a. }\, \ln^2\left(2x\, -\, 1\right)\, =\, 4$ . . . . .$\mbox{b. }\, \ln\left({\left(2x\, -\, 1\right)^2}\right)\, =\, 4$ RudyTheRocka wrote:5. logbase=2(12) The expression, if I understand you correctly, is as follows: . . . . .$\log_2{(12)}$ But what are you supposed to do with this? What were the instructions? RudyTheRocka wrote:6. The amount of money in a bank account can be expressed by the exponential equation A=300(1.005)^12(t) where A is the amount in dollars and t is the time in years. About how many years will it take for the amount in the account to be more than$900.

You are given a formula relating A and t. You are given a value for A, and are asked to find t. So plug the given value in for the given variable, and solve the resulting exponential equation for t.

Start by dividing through by the 300. Then take the log of each side, use log rules to move the "12t" down in front of the "logwhatever base(1.005)", and then divide off the log and the 12 to isolate t (that is, to get t by itself on one side of the "equals" sign).

Evaluate the resulting expression in your calculator to get a decimal approximation for your answer. Remember to put an appropriate unit (in this case, "years") on your final answer.

stapel_eliz

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