How do you evaluate ${ln e}^{8} - 7 {ln e}^{3}$ $\ln e ^ { 8} - 7\ln e ^ { 3}$ ?

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Answer 1 · 2018-04-11T12:05:14

$13$ $13$

logarithm rule:

${log a}^{n} = n log a$ $log a^n = n log a$

this also applies for ${log}_{e} : {ln a}^{n} = n ln a$ $log_e: ln a^n = n ln a$ .

using this rule, ${ln e}^{8}$ $ln e^8$ is the same as $8 ln e$ $8 ln e$ .

$ln e$ $ln e$ is the power that $e$ $e$ is raised by to equal $e$ $e$ . this power is $1$ $1$ .

$ln e$ $ln e$ is $1$ $1$ , so $8 ln e$ $8 ln e$ is $8$ $8$ .

${ln e}^{3}$ $ln e^3$ is the same as $3 ln e$ $3 ln e$ .

$7 {ln e}^{3}$ $7 ln e^3$ is the same as $7 \cdot 3 ln e$ $7 * 3 ln e$ .

$7 \cdot 3 = 21$ $7 * 3 = 21$ , so $7 {ln e}^{3}$ $7 ln e^3$ is $21 ln e$ $21 ln e$ .

$(8 ln e) - (21 ln e)$ $(8 ln e) - (21 ln e)$ is $8 - 21$ $8 - 21$ .

$8 - 21 = - 13$ $8 - 21 = -13$

hence, ${ln e}^{8} - 7 {ln e}^{3} = 8 - 21 = - 13$ $ln e^8 - 7 ln e^3 = 8 - 21 = -13$ .

How do you evaluate lne8−7lne3\ln e ^ { 8} - 7\ln e ^ { 3} ?