Hi guys,
how to solve this: 1/3 log(base5) 3x + log(base5) 4=3
Please help math hero
Please clarify parentheses
1/3 log(base5) 3x + log(base5) 4=3
perhaps you mean:
1 / [3 log(base5) 3x + log(base5) 4 ] = 3
1 / [ log(base5) 27x^3 + log(base5) 4]=3
1 / log(base5) 108 x^3 =3
1 = 3 log(base5) 108 x^3
log(base5) 108^3 x^9 = 1
so 108^3 x ^9 = 5
about 0.25 ?
1/3log(base5) 3x + log(base5) 4=3
what I meant is: 1/3=one-third ,log5= log(base5)
assuming base 5 throughout,
1/3 log3x + log4 = 3
log3x + 3log4 = 9
log(3x * 4^3) = 9
192x = 5^9
x = 5^9/192
Hello! I can help you with that math problem.
To solve the equation 1/3 log(base5) (3x) + log(base5) 4 = 3, we can use logarithmic properties to simplify the equation.
First, let's combine the two logarithms using the product rule of logarithms, which states that log(base a) M + log(base a) N = log(base a) (M * N). Applying this to our equation, we have:
log(base5) (3x)^(1/3) * 4 = 3
Next, let's convert the equation into exponential form. The logarithmic expression log(base a) M = N is equivalent to the exponential expression a^N = M. Applying this to our equation, we have:
(3x)^(1/3) * 4 = 5^3
Now, we can simplify further. We can remove the exponent by raising both sides of the equation to the power of 3. After doing this, we get:
(3x) * 4^3 = 5^3
Simplifying further, we have:
3x * 64 = 125
Now, we can solve for x. Divide both sides of the equation by 3:
3x = 125 / 64
Finally, divide both sides of the equation by 3 to isolate x:
x = (125 / 64) / 3
So, the solution to the equation is x = 125 / (64 * 3), or x = 125 / 192.
I hope that helps! Let me know if you have any further questions.