solving for x exponential form...need help!

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sportsaholic2397

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Jun 12, 2009
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alright so i need help with this... i think where i am getting thrown off is there is an x to be solved for on both sides of the equation but here it is. please help.

solve for x:

10^(2x-1)=e^(4x-3)
 


Hello sportsaholic2397:

One way to start is to take the natural logarithm of both sides.

Then use the logarithmic property that allows moving the expressions 2x-1 and 4x-3 out of the exponent position.

You'll get the following linear equation that can be solved in the usual way.

(2x - 1) * ln(10) = 4x - 3

If you would like any more help, then please show whatever work that you can or explain what you're thinking, so that we might determine where to continue helping you.

Cheers,

~ Mark

 
so I understand what you have done so far but I plugged in the one x value i came up with which was 1 but it did not plug in to both x's and give me the same answer...so are there two different x values and if there are i am unsure of how to find them starting from the linear equation we (or you) have already come up with please get back to me a.s.a.p
thank you
 


Hey there:

This equation has only one solution. Obviously, I can't tell how you obtained x = 1, but that's not correct.

The natural logarithm of 10 is a constant. We can continue to use the symbolism ln(10) to represent this irrational number throughout our algebraic manipulations, while solving for x.

This way, we'll end up with an expression for x that represents the exact solution.

As the very last step, we can use a scientific calculator to write some decimal approximation for the solution.

So, you understand how to arrive at the following equation, right?

(2x - 1) * ln(10) = 4x - 3

The next step is to use the Distributive Property to multiply the expression (2x - 1) by the number ln(10), on the lefthand side.

2x * ln(10) - ln(10) = 4x - 3

Did you do this?

Again, ln(10) is a constant. So is -3. They are like terms. Use algebra to get all of the constants on one side of the equation, and then combine them.

2x and 4x are like terms. Use algebra to get both of them on the other side of the equation, and then combine them.

Does this make sense? If not, then please reply with specific questions or show your work. We can't point out your erroneous ways, unless we first see them. 8-)

Cheers,

~ Mark

 
alright so i understand getting constants all on one side, and like terms on the other. now i noticed that you have said to do 2x*ln(10) -ln(10) =4x-3
and then asked if i did that. i did not. now with constants you are suppose to subtract from both sides of the equation correct and i see you put ln(10)-ln(10), do you subtract ln(10) from -3 as well? and then i combine the x's which i get..
 
sportsaholic2397 said:
… i see you put ln(10)-ln(10) …
Click to expand...


We do not have ln(10) - ln(10).

We have 2x * ln(10) - ln(10).

Are you familiar with the concept known as Order of Operations? It tells us that we cannot subtract the term ln[10] from the factor ln(10) in the product 2x*ln(10).

Continue with the following equation and steps.

2x * ln(10) - ln(10) = 4x - 3

Step 1: Add ln(10) to both sides

Step 2: Subtract 4x from both sides

Step 3: Factor x out from the lefthand side

Step 4: Divide both sides by the other factor, to solve for x

8-)

 
alright so i add ln(10) to both sides and then i subtracted 4x from both sides, and i have -2x*4.60(or keep as ln10?)=-.69 .....then you say factor x out from the left hand side...and thats where i got confused...i dont understand that step because i thought you already eliminate the x on the right side by subtracting the 4x from both sides and so that leaves you with just -2x which i thought you made into x when you divide that by the other piece..
 
sportsaholic2397 said:
… i subtracted 4x from both sides, and i have -2x*4.60(or keep as ln10?)=-.69
Click to expand...


I'm getting an impression that you're not really paying attention to me.

We do not have 2x - 4x.

We have 2x * ln(10) - 4x.

You need to learn the prerequisite algebra, before attempting to solve linear equations. We cannot do the subtraction 2x-4x on the lefthand side because of something VERY IMPORTANT that's known as Order of Operations.

I also already answered your question above, regarding whether or not to wait until the very end to evaluate the expression ln(10). Do you remember reading it?

If I wrote anything that you do not understand, then the appropriate rejoinder is to ask specific questions.

I instructed you to factor x out of the expression 2x * ln(10) - 4x.

Can you do this?

 
Mr Sportsman, Mark asked you to factor this:
"I instructed you to factor x out of the expression 2x * ln(10) - 4x"
HINT:
7x * k - 5x would be this way: x(7k - 5)

By the way, your posts are kinda sloppy (hard to follow, poor grammar);
why don't you get more serious? Look at Mark's posts: can you start using his "clarity" style?

Would be much easier for you to get help if you did.

[You're welcome, Mark!]
 
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