SOLUTION: "can u pls. help me write the following exponential equations in logarithmic form?" 1.7^x=49 2.6^-3=(1/126) 3.12^2=144 also this...find the value of the following logarithmic exp

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: "can u pls. help me write the following exponential equations in logarithmic form?" 1.7^x=49 2.6^-3=(1/126) 3.12^2=144 also this...find the value of the following logarithmic exp      Log On


   

Question 1187994: "can u pls. help me write the following exponential equations in logarithmic form?" 1.7^x=49 2.6^-3=(1/126) 3.12^2=144 also this...find the value of the following logarithmic expressions. 1.log3 243 2.log6 (1/216) 3.log0.25 16 thank u so much for the help.
Found 3 solutions by MathLover1, MathTherapy, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
1.
1.7%5Ex=49+

log%281.7%5Ex%29=log%2849%29+

x%2Alog%281.7%29=log%2849%29+

x%2Alog%2817%2F10%29=log%287%5E2%29+

x%2A%28log%2817%29-log%2810%29%29=2log%287%29+

x=2log%287%29+%2F%28log%2817%29-log%2810%29%29.......log%2810%29=1

x=2log%287%29+%2F%28log%2817%29-1%29

x≈7.33436


2.

2.6%5E-3=%281%2F126%29+=> assuming you have 2.6%5E%28-3x%29=%281%2F126%29+

2.6%5E%28-3x%29=%281%2F126%29+.............2.6%5E-3=%2826%2F10%29%5E-3=%2813%2F5%29%5E-3

%2813%2F5%29%5E%28-3x%29=%281%2F126%29+

log%28%2813%2F5%29%5E%28-3x%29%29=log%28%281%2F126%29%29+

-3x%2Alog%2813%2F5%29=log%28%281%2F126%29%29+

-3x%28log%2813%29-log%285%29%29=log%281%29-log%28126%29+........log%281%29=0

-3x%28log%2813%29-log%285%29%29=-log%28126%29+..............2+log%283%29+%2B+log%2814%29+

-3x%28log%2813%29-log%285%29%29=-%282+log%283%29+%2B+log%2814%29%29+

x=-%282+log%283%29+%2B+log%2814%29%29%2F%28-3%28log%2813%29-log%285%29%29%29+

x=-%282+log%283%29+%2B+log%2814%29%29%2F%28-3%28log%2813%29-log%285%29%29%29+

x≈1.68715

3.

3.12%5E2=144 => assuming you have 3.12%5E%282x%29=144

log%283.12%5E%282x%29%29=log%28144%29

%282x%29log%283.12%29=log%2812%5E2%29

%282x%29=%282log%2812%29%29%2Flog%28312%2F100%29

%282x%29=%282log%283%2A4%29%29%2F%28log%28312%29-log%28100%29%29............log%28100%29=2, write 312=2%5E3%2A3%2A13
2x=%282log%283%29%2Blog%284%29%29%2F%28log%282%5E3%2A3%2A13%29-2%29





x+=+2.18389



1.
log%283+%2C243%29=log%283%5E5%29%2Flog%283%29=+5log%283%29%2Flog%283%29=5
2.

3.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

"can u pls. help me write the following exponential equations in logarithmic form?" 1.7^x=49 2.6^-3=(1/126) 3.12^2=144 also this...find the value of the following logarithmic expressions. 1.log3 243 2.log6 (1/216) 3.log0.25 16 thank u so much for the help. 
This is in no way as COMPLEX as that woman made it out to be.


If you need a value, then just enter that into a calculator. 
If it's difficult to enter, then just enter the change-of-base version: log+%28%2849%29%29%2Flog+%28%2817%29%29
You can now do nos. 2 & 3 the same way.


The value of the following logarithmic expression: highlight_green%28matrix%281%2C2%2C+%221.%22%2C+log+%283%2C+243%29%29%29
Enter the above into a calculator. 
Again, if it's difficult to do so, then just enter the change-of-base version: log+%28%28243%29%29%2Flog+%28%283%29%29
If you need to calculate without a calculator, then do the following:
 matrix%281%2C3%2C+x%2C+%22=%22%2C+log+%283%2C+%28243%29%29%29 ----- Setting log+%283%2C+%28243%29%29 equal to x
matrix%281%2C3%2C+3%5Ex%2C+%22=%22%2C+243%29 ----- Converting to EXPONENTIAL form
matrix%281%2C3%2C+3%5Ex%2C+%22=%22%2C+3%5E5%29 ------ Converting 243 to base 3
  ------ Bases are equal and so are their exponents

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I suspect the way you posted the problems misled the other tutors. Very probably the first set of three problems are these:

(1) 7%5Ex=49

(2) 6%5E%28-3%29=1%2F216 <-- that one includes a typo -- you showed 1/126....

(3) 12%5E2=144

To convert each of those to logarithmic form, all you need to remember is this:

THE LOGARITHM IS THE EXPONENT

(1) The exponent is x; obviously the base is 7 --> "log (base 7) of 49 is x" --> log%287%2C%2849%29%29=x

(2) The exponent is -3; the base is 6 --> "log (base 6) of 1/216 is -3" --> log%286%2C%281%2F216%29%29=-3

(3) The exponent is 2; the base is 12 --> "log (base 12) of 144 is 2" --> log%2812%2C%28144%29%29=2

For the other problems you again only need to remember the same thing.

(1) log%283%2C%28243%29%29=x --> "log (base 3) of 243 is x" --> THE LOGARITHM (x) IS THE EXPONENT --> 3%5Ex=243

(2) log%286%2C%281%2F216%29%29=x --> "log (base 6) of 1/216 is x" --> THE LOGARITHM (x) IS THE EXPONENT --> 6%5Ex=1%2F216

(3) log%280.25%2C%2816%29%29=x --> "log base 0.25 of 16 is x" --> THE LOGARITHM (x) IS THE EXPONENT --> 0.25%5Ex=16

I leave it to you to find the values of x for those last three problems.