Final answer:
Using the properties of exponents and logarithms, including the product rule for logarithms, we proved that log_mn is equal to the sum of log_m and log_en by expressing the number n as e raised to the power of ln(n) and applying logarithmic rules.
Step-by-step explanation:
To prove that log_mn = log_m + log_en, we can utilize logarithmic properties. Given that log_mn=y and log_gn=y, we want to demonstrate that the logarithm of n with respect to base m can be expressed as the sum of the logarithm of m and the natural logarithm of n. Considering the properties of exponents and logarithms, specifically the property that states the logarithm of a product is the sum of the logarithms (log xy = log x + log y), we can conclude the following:
- log_mn is the same as log_m + log_n for base m.
- To incorporate the natural logarithm (ln), remember that any number a can be expressed as e raised to ln(a). So we rewrite n as e raised to the power of ln(n).
- Now, we can use the property that the logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number (log_b(a^x) = x*log_b(a)).
- Hence, log_m(e^ln(n)) = ln(n)*log_m(e), with log_m(e) being a constant.
- Because log_mn = log_m + ln(n), the proof is complete.
Therefore, with the use of logarithmic and exponential properties, we have proven the required expression for logarithms.