logarithm question?

Question

Hey everyone we are currently studying logarithms in grade 12 advanced functions and I am having a problem with this question:

4.6 * 1.06^(2x + 3) = 5 * 3^x [Ans: 0.09]

How would you solve for x? The answer is probably simple but I'm just not seeing it right now. Any help?

Thanks.

Answer 1

Take the log of both sides:
log( 4.6 * 1.06^(2x + 3) ) = log( 5 * 3^x )

Remember log(ab) = log(a) + log(b)
log( 4.6 ) + log(1.06^(2x + 3)) = log( 5 ) + log(3^x)

Next remember that log(a^k) = k log(a)
log( 4.6) + (2x + 3) log(1.06) = log( 5) + x log(3)
log (4.6) + 2x log(1.06) + 3 log(1.06) = log(5) + x log(3)

Let's get all the x terms on one side:
2 log(1.06) x - log(3) x = log(5) - 3 log(1.06) - log(4.6)

Distribute out an x:
x (2 log(1.06) - log(3)) = log(5) - 3 log(1.06) - log(4.6)

Now divide:
x = ( log(5) - 3 log(1.06) - log(4.6) ) / (2 log(1.06) - log(3))

According to Google calculator (link below):
x ≈ 0.0930938722

As a double check:
Left side: 4.6 * 1.06^(2(0.0930938722) + 3) ≈ 5.53843508
Right side: 5 * (3^0.0930938722) ≈ 5.53843508

Looks good to me... unless you want to round the answer a little.

x ≈ 0.0931

Answer 2

You need to already know that

#1) log [b^x] = x log b; and #2) log [ab] = (log a) + (log b)

Taking the log of each side gives

log 4.6 + [ (2x + 3) log 1.06] = log 5 + x[log 3]

Simplify by mult within left brace; get all terms with x to right, (get other terms, that don't have x as a factor to left) and factor out x gives:

log 4.6 + 3 log 1.06 - log 5 = x [log 3 - 2 log 1.06] then divide each side by the right brace quantity to get x.
--------------
My computer's calculator gives this answer:
0.093093872206630496713132347620487 which you prolly should ignore after the first significant digit.

Answer 3

4.6 * 1.06^(2x + 3) = 5 * 3^x [Ans: 0.09]
(4.6 )1.06^(2x + 3) = 5 * 3^x [Ans: 0.09] extract log of both sides
(2x+3) log 1.06 +log 4.6 = x log 3 + log 5
(2x+3) log 1.06 - x log 3 = log 5 - log 4.6
2x log 1.06 + 3 log 1.06 - x log 3 = log 5 - log 4.6
2x log 1.06 - x log 3 = log 5 - log 4.6 - 3 log 1.06
2x (0.0253) - x (0.4771212) = 0.699 - 0.66276 - 3(0.0253)
0.5277212 x = - 0.03971
x = - 0.075248

Answer 4

4.6*1.06^(2x+3)=5*3^x
1.06^(2x+3)=(5/4.6)*3^x < log(base1.06) of ((5/4.6)*3^x)=2x+3
Now we'll use the rule log(a*b)=loga+logb
log<1.06>of (5/4.6) + log<1.06>of3^x=2x+3
log<1.06>(5/4.6)+x*log<1.06>3=2x+3 < log<1.06>(5/4.6)=2x+3-x*log<1.06>(3)
log<1.06>(5/4.6)-3=(2-log<1.06>(3))x
(log<1.06>(5/4.6)-3)/(2-log<1.06>(3))=x
There it is...
"Logarithm of (5/4.6) with base 1.06 minus 3, over 2 minus logarithm of 3 with base 1.06 is equal to X"

Answer 5

take log..

log 4.6 + (2x+3)log1.06 = log 5 + xlog3
0.663 + (2x+3)0.025 = 0.699 + x0.477
0.05x +0.738 = 0.699 + 0.477x
0.427x = 0.039
x = 0.09

Answer 6

4.6 * 1.06^(2x+3) = 5* 3^x

take logs

log(4.6)(1.06)^(2x+3) = log(5)(3^x)

apply log rules log(ab) = loga + logb and log(a^b) = blog(a)

log(4.6) + log(1.06)^(2x+3) = log(5) + log(3^x)

simplify

log(4.6) + (2x+3)log(1.06) = log(5) + x log(3)

log(4.6) + 2xlog(1.06) + 3log(1.06) = log(5) + x log(3)

2xlog(1.06) - x log(3) = log(5) - log(4.6) - 3log(1.06)

=>x(0.116) - x (1.1) = 1.61 - 1.53 - 0.174

=>- 0.984x = -0.094

x = 0.094/0.984 = 0.095

Answer 7

Solve for x with logs.
4.6*1.06^(2x + 3) = 5*3^x
______________

Remember the rules of logs.

log(ab) = log(a) + log(b)
log(a^b) = b*log(a)
___________

4.6*1.06^(2x + 3) = 5*3^x

Take the natural log of both sides.

ln[4.6*1.06^(2x + 3)] = ln[5*3^x]
ln(4.6) + (2x + 3)ln(1.06) = ln(5) + x*ln(3)
ln(4.6) + (2x)ln(1.06) + 3ln(1.06) = ln(5) + x*ln(3)
ln(4.6) + x*ln(1.06²) + ln(1.06³) = ln(5) + x*ln(3)

Gather the x terms to one side.

x*ln(1.06²) - x*ln(3) = ln(5) - ln(4.6) - ln(1.06³)
x*ln(1.06²/3) = ln[5/(4.6*1.06³)]

x = ln[5/(4.6*1.06³)] / ln(1.06²/3)
x ≈ 0.0930938

Answer 8

hint: opposite powers with the aid of using logarithms. So opposite a^n with the aid of taking the log of a^n with a base a; n=log[a](a^n) And opposite log base a with the aid of taking the potential of a. a^(log[a](n))=n.