How do you solve ${log}_{5} (5) + {log}_{5} (2 x) + {log}_{5} (3) = 98$ $log_5(5)+log_5(2x)+log_5(3)=98$ ?

Question

How do you solve ${log}_{5} (5) + {log}_{5} (2 x) + {log}_{5} (3) = 98$ $log_5(5)+log_5(2x)+log_5(3)=98$ ?

1 Answer

Impact of this question

1922 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-05T05:20:38

Step by step explanation is given below.

Explanation:

Before starting out on the problem let us quickly go through few rules.

$1 . {log}_{b} (a) = k \Rightarrow a = b^{k}$ $1. log_b(a) = k => a=b^k$ this is used to covert logarithm into exponent form.
$2, {log}_{b} (b) = 1$ $2, log_b(b) = 1$
$3 . {log}_{b} (P) + {log}_{b} (Q) = {log}_{b} (P Q)$ $3. log_b(P) + log_b(Q) = log_b(PQ)$ This rule for condensing the log sum to product.

For our problem, these rules are enough. For the other type of logarithmic equations, please do spend some time learning all the basic rules.

${log}_{5} (5) + {log}_{5} (2 x) + {log}_{5} (3) = 98$ $log_5(5) +log_5(2x)+log_5(3) = 98$

$1 + {log}_{5} ((2 x) (3)) = 98$ $1 + log_5((2x)(3))=98$

Subtract $1$ $1$ from both the sides

${log}_{5} (6 x) = 97$ $log_5(6x)=97$

$6 x = 5^{97}$ $6x = 5^(97)$

Dividing by $6$ $6$ on both the sides to solve for $x$ $x$

$x = \frac{5^{97}}{6}$ $x=5^(97)/6$ Answer

Science

Math

Humanities

... and beyond

How do you solve ${log}_{5} (5) + {log}_{5} (2 x) + {log}_{5} (3) = 98$ $log_5(5)+log_5(2x)+log_5(3)=98$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve log5(5)+log5(2x)+log5(3)=98log_5(5)+log_5(2x)+log_5(3)=98 ?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve ${log}_{5} (5) + {log}_{5} (2 x) + {log}_{5} (3) = 98$ $log_5(5)+log_5(2x)+log_5(3)=98$ ?