Find the value of r, if log r + log r2 +log r4 +log r8+log r16 +log r32 =63?
I will assume that your log r + log r2 +log r4 +log r8+log r16 +log r32 =63
and the base of the log is 10
is actually log r + log r^2 +log r^4 + .... + log r^32 =63
then logr + 2logr + 4logr + .... 32logr = 63
logr( 1 + 2 + 4 + ... + 32) = 63 , 6 terms inside the bracket
this is a geometric series, a = 1, r = 2, n = 6
sum(6) = 1(2^6 - 1)/(2-1) = 63 , or we could have just added them
63 logr = 63
logr = 1
r = 10
To find the value of r in the equation:
log r + log r^2 + log r^4 + log r^8 + log r^16 + log r^32 = 63
We can use the properties of logarithms to simplify the equation:
Since the sum of logarithms is equal to the logarithm of the product, we can write the equation as:
log (r * r^2 * r^4 * r^8 * r^16 * r^32) = 63
Simplifying further, we have:
log r^(1 + 2 + 4 + 8 + 16 + 32) = 63
log r^63 = 63
Using the property that if log base a of x is equal to y, then a^y = x, we have:
r^63 = 10^63
Taking the 63rd root of both sides, we have:
r = (10^63)^(1/63)
Simplifying further, we get:
r = 10^(63/63)
Since any number raised to the power of 1 is equal to the number itself, we have:
r = 10
The value of r is 10.
To find the value of r in the equation log r + log r^2 +log r^4 + log r^8 + log r^16 + log r^32 = 63, we can start by using the logarithmic property that states that log a + log b = log (a * b).
Applying this property, we can simplify the equation as follows:
log r + log r^2 + log r^4 + log r^8 + log r^16 + log r^32 = 63
log (r * r^2 * r^4 * r^8 * r^16 * r^32) = 63
log (r^(1 + 2 + 4 + 8 + 16 + 32)) = 63
log (r^63) = 63
Therefore, we have r^63 = 10^63 (using the fact that the logarithm base 10 is commonly used)
To take the logarithm of both sides, with the same base, we get:
63 log r = 63
Now, we can divide both sides by 63 to isolate log r:
log r = 1
To find the value of r, we need to find the antilogarithm (exponentiation) of both sides:
r = 10^(log r)
Since log r is equal to 1, we have:
r = 10^1
Therefore, the value of r is 10.