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February 28, 2015

February 28, 2015

Posted by **Anonymous** on Monday, September 20, 2010 at 9:21pm.

Find the exact alue of x for which (4^x)*(5^[4x+3])=(10^[2x+3])

I can't seem to come to a solution.

We're reviewing last year's lessons, so change of base and logarithmic expressions are what we're going over right now.

Here's what I've done so far:

(4^x)*(5^[4x+3])=(10^[2x+3])

log both sides

log[(4^x)*(5^[4x+3])]=log[(10^[2x+3])]

(xlog4)+([4x+3]log5)=([2x+3]log10)

since log10=1

(xlog4)+([4x+3]log5)=(2x+3)

Have I started correctly? Where do I go from here? Please be detailed, I want to understand this.

- Math -
**MathMate**, Monday, September 20, 2010 at 9:36pmProceed and solve for x in terms of ln.

I get x=ln(8)/ln(25).

- Math -
**Anonymous**, Monday, September 20, 2010 at 9:41pmHow do I proceed? I really have no clue where to go from where I've stopped! Usually I can simply factor out an x from the LS, but here I can't do that. Could you please show me your solution with every step? It would be so helpful.

Thank you in advance :)

- Math -
**MathMate**, Monday, September 20, 2010 at 9:51pmIt is a linear equation with numerical coefficients (log4, etc.).

Expand

(xlog4)+([4x+3]log5)=(2x+3)

to give

xlog4 + 4xlog5 - 2x = 3 -3log5

Factor out x on the LHS and solve for x.

All the log4, log5 are numerical values that you can simplify eventually.

- Math -
**Anonymous**, Monday, September 20, 2010 at 10:02pmSorry to keep bugging, but I followed what you told me and got to:

x=(3-[log125])/([log2500]-2)

How can I get to the answer from that? I don't know how to deal with the 3 and the -2 that are still lying around.

- Math -
**Anonymous**, Monday, September 20, 2010 at 10:09pmAlright, what I did now was:

since log10=1, log100=2, and log1000=3

So I plugged that in to what I had

x=(log1000-log125)/(log2500-log100)

x=(log[1000/125])/(log[2500/100])

x=(log8)/(log25)

Is that what I was supposed to do?

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