Given that loga7 = 0.845, and loga2 = 0.301, find loga343
A 0.603
B 2.535
C 1.146
D 0.254
343 = 7^3
loga(343) = 3 loga(7)
Assuming you mean:
Given that loga(7) = 0.845, and loga(2) = 0.301, find loga(343).
Using the rule of logarithm:
loga(xn) = n*loga(x) .............(1)
From 7^3=343, we substitute in (1) x=7, n=3, so
loga(343)
=loga(73)
= 3*loga(7)
= 3*0.845
= ?
To find loga343, we need to express 343 as a power of 7.
Since 7^3 = 343, we can write loga343 as loga(7^3).
Using the logarithmic identity loga(b^c) = c * loga(b), we can express loga(7^3) as 3 * loga(7).
Given that loga7 = 0.845, we can substitute it into the expression to find loga343:
loga343 = 3 * loga7 = 3 * 0.845 = 2.535
Therefore, the answer is option B: 2.535.
To find the value of log base a of 343, we can use the property of logarithms that states that the logarithm of a power is equal to the product of the exponent and the logarithm of the base number.
In this case, since 343 is equal to 7 raised to the power of 3 (7^3 = 343), we can rewrite log base a of 343 as 3 * log base a of 7.
Given that log base a of 7 is 0.845, we can substitute this into the equation:
log base a of 343 = 3 * log base a of 7 = 3 * 0.845 = 2.535
Therefore, the value of log base a of 343 is 2.535.
Hence, the correct option is B.