How do you solve $2^x*5=10^x$ ?

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Answer 1 · 2016-06-30T00:53:32

If $A = B$ , then $logA = logB$ :

$log(2^x xx 5) = log(10^x)$

Use the rule $log_a(n/m) = log_an - log_am$ to simplify.

$log2^x + log5 = log10^x$

Use the rule $loga^n = nloga$ to simplify further.

$xlog2 + log5 = xlog10$

$log5 = xlog10 - xlog2$

Factor out the $x$ .

$log5 = x(log10 - log2)$

Put back the logs on the left side into quotient form to simplify.

$log5 = x(log(10/2))$

$log5 = x(log5)$

Isolate x.

$log5/log5 = x$

$1 = x$

Checking in the original equation, we find that this solution works.

Hopefully this helps!

Answer 2 · 2016-06-30T00:56:28

$x=1$ (using alternate solution method)

If $2^x*5=10^x$

$rArr 2^x*5 =(2*5)^x$

$rArr 2^x*5 = 2^x*5^x$

$rArr 5=5^x$

$rArr x=1$

How do you solve 2^x*5=10^x2^x*5=10^x?